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2019 - 1901(13) - 1902(11)

Any replacements are listed farther down

[1945] **viXra:1902.0332 [pdf]**
*submitted on 2019-02-21 03:55:20*

**Authors:** Toshiro Takami

**Comments:** 5 Pages.

product [prime (n) ^ s / (prime (n) ^ s - 1), {n, 39001, 40000}], [s = 0.5 + i 14.1347]
Because there was capacity limitation of the computer, I divided it up to 40000, but there was no convergence trend.
This seems to be a conspiracy by ζ stars.

**Category:** Number Theory

[1944] **viXra:1902.0235 [pdf]**
*submitted on 2019-02-13 05:23:19*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 6 Pages. Submitted to the The Ramanujan Journal. Comments welcome.

In this paper, we consider the abc conjecture in the case c=a+1. Firstly, we give the proof of the first conjecture that c

**Category:** Number Theory

[1943] **viXra:1902.0200 [pdf]**
*submitted on 2019-02-11 06:24:07*

**Authors:** Faisal Amin Yassein Abdelmohssin

**Comments:** 2 Pages.

I claim that the sum of following distinct proper fractions [(1/2),(1/3),(1/6)] is the only triple of distinct proper fraction that sum to 1 {i.e. [(1/2)+(1/3)+(1/6)]=1}.

**Category:** Number Theory

[1942] **viXra:1902.0147 [pdf]**
*submitted on 2019-02-08 09:11:21*

**Authors:** Kenneth A. Watanabe

**Comments:** 13 Pages.

The Near-Square Prime conjecture, states that there are an infinite number of prime numbers of the form x^2 + 1. In this paper, a function was derived that determines the number of prime numbers of the form x^2 + 1 that are less than n^2 + 1 for large values of n. Then by mathematical induction, it is proven that as the value of n goes to infinity, the function goes to infinity, thus proving the Near-Square Prime conjecture.

**Category:** Number Theory

[1941] **viXra:1902.0117 [pdf]**
*submitted on 2019-02-06 21:33:23*

**Authors:** Toshiro Takami

**Comments:** 3 Pages.

In the mathematical world there was a crawl of analytical connections walking around and I noticed the fact that the mathematics world was dominated by the monster.
You should wake up quickly escape from the brainwashing analysis connection (monster).
I notice this, I will dare to write for justice.
I have tried to prove Riemann hypothesis, I noticed this.
That is, Riemann hypohesis is fundamentally completely mistake.

**Category:** Number Theory

[1940] **viXra:1902.0106 [pdf]**
*submitted on 2019-02-06 07:50:18*

**Authors:** Algirdas Anatano Maknickas

**Comments:** 2 Pages.

This remark gives analytical solution of Last Fermat's Theorem

**Category:** Number Theory

[1939] **viXra:1902.0040 [pdf]**
*submitted on 2019-02-02 16:34:38*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 9 Pages. A Complete proof of the abc conjecture using elementary calculus with numerical examples. Submitted to the Ramanujan Journal. Your comments are welcome.

In this paper, we consider the abc conjecture. Firstly, we give a proof of a first conjecture that c

**Category:** Number Theory

[1938] **viXra:1902.0036 [pdf]**
*submitted on 2019-02-03 00:15:01*

**Authors:** Simon Plouffe

**Comments:** 10 Pages.

Une famille de formules permettant d'obtenir une suite de longueur arbitraire de nombres premiers. Ces formules sont nettement plus efficaces que celles de Mills ou Wright. Le procédé permet de produire par exemple une suite dont la croissance est double exponentielle mais l'exposant = 101/100.

**Category:** Number Theory

[1937] **viXra:1902.0005 [pdf]**
*submitted on 2019-02-01 10:53:12*

**Authors:** James Edwin Rock

**Comments:** 2 Pages of exposition. 5 Tables. Copyright 2019 James Edwin Rock Creative Commons Attribution-ShareAlike 4.0 International License

Let P_n be the n_th prime. For twin primes P_n – P_(n-1) = 2. Let X be the number of (6j –1, 6j+1) pairs in the interval [P_n, P_n^2]. The number of twin primes (TPAn) in [P_n, P_n^2] can be approximated by the formula
(a_3 /5)(a_4 /7)(a_5 /11)…(a_n /P_n)(X) for 3 ≤ m ≤ n, a_m = P_m –2 .
We establish a lower bound for TPAn (3/5)(5/7)(7/9)…(P_n–2)/P_n)(X) = 3X/P_n < TPAn.
We exhibit a formula showing as P_n increases, the number of twin primes in the interval [P_n, P_n^2] also increases. Let P_n – P_(n-1) = c. For all n (TPAn-1)(1+(2c –2)/2P_(n-1)+(c^2–2c)/2P_(n-1)^2) < TPAn

**Category:** Number Theory

[1936] **viXra:1901.0436 [pdf]**
*submitted on 2019-01-29 10:25:44*

**Authors:** Kenneth A. Watanabe

**Comments:** 10 Pages.

Legendre's conjecture, states that there is a prime number between n^2 and (n + 1)^2 for every positive integer n. In this paper, an equation was derived that determines the number of prime numbers less than n for large values of n. Then by mathematical induction, it is proven that there is at least 1 prime number between n^2 and (n + 1)^2 for all positive integers n thus proving Legendre’s conjecture.

**Category:** Number Theory

[1935] **viXra:1901.0430 [pdf]**
*submitted on 2019-01-29 01:24:31*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we consider the $ABC$ conjecture,then we give a proof that $C<rad^2(ABC)$ that it will be the key of the proof of the $ABC$ conjecture.

**Category:** Number Theory

[1934] **viXra:1901.0300 [pdf]**
*submitted on 2019-01-19 14:41:09*

**Authors:** Toshiro Takami

**Comments:** 4 Pages.

On calculation by Euler 's formula(3), at least notice that the zero point of the real part is not on x = 0.5.
Using Euler 's formula, we found that at least the real part' s zero point is not x = 0.5 but about x = 0.115444. Moreover, the imaginary point is around i14.524.
And s=0.8355 +i39.
And s=0.1645 +i39.
And s=0.884556 +i14.524.
And s=0.115444 +i14.524
Also, replacing sin with cos, the imaginary part becomes zero.
I do not know at all whether the collapse of Riemann hypothesis or not?
In addition, books are printed as cos instead of sin.
Also, I have collected ζ on the left side.
In addition, (8) is Euler's formula found overseas, which is also a singular point in this. Also, I have collected ζ on the left side.
In this case, sin is printed instead of cos, but if sin and cos are exchanged, the zero point moves only from the real part to the imaginary part.

**Category:** Number Theory

[1933] **viXra:1901.0297 [pdf]**
*submitted on 2019-01-19 22:35:49*

**Authors:** Michael Grützmann

**Comments:** 2 Pages.

every prime number can be a sum of p=3+...+3+2 or
q=3+...+3+4, the number of '3's in both equatons always being odd.
there are the for two primes p+q the combinations
'p+q', 'p+p' and 'q+q'.
we consider case 1: p+q=3k+2+3l+4
3k must be odd, as the product of two odd numbers,
3l must be odd, for the same reason.
but the sum of two odd numbers is an even number always. also, if you add more even numbers, like 2 and 4, the result will always be even also.
So this results in an even number.
cases 'p+p' and 'q+q' analogue.

**Category:** Number Theory

[1932] **viXra:1901.0227 [pdf]**
*submitted on 2019-01-16 12:07:43*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License

Collatz sequences are formed by applying the Collatz algorithm to any positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. If the Collatz conjecture is true eventually you always get back to one. A connected Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[1931] **viXra:1901.0155 [pdf]**
*submitted on 2019-01-11 06:22:54*

**Authors:** Edgar Valdebenito

**Comments:** 1 Page.

This note presents three trigonometric identities.

**Category:** Number Theory

[1930] **viXra:1901.0116 [pdf]**
*submitted on 2019-01-10 02:36:15*

**Authors:** Quang Nguyen Van

**Comments:** 2 Pages.

The equation a^5 + b^5 = c^2 has no solution in integer. However, related to Fermat- Catalan conjecture, the equation a^5 + b^5 = 2c^2 has a solution in integer. In this article, we give a parametric equation of the equation a^5 + b^5 = 2c^2.

**Category:** Number Theory

[1929] **viXra:1901.0108 [pdf]**
*submitted on 2019-01-08 11:13:12*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. A Proof of ABC conjecture is submitted to the Journal of Number Theory (2019). Comments Welcome.

In this paper, we assume that the ABC conjecture is true, then we give a proof that Beal conjecture is true. We consider that Beal conjecture is false then we arrive to a contradiction. We deduce that the Beal conjecture is true.

**Category:** Number Theory

[1928] **viXra:1901.0104 [pdf]**
*submitted on 2019-01-08 18:01:42*

**Authors:** Toshiro Takami

**Comments:** 5 Pages.

I tried to prove that(Riemann hypothesis), but I realized that I can not prove how I did it.
When we calculate by the sum method of (1) we found that the nontrivial zero point will never converge to zero.
Calculating ζ(2), ζ(3), ζ(4), ζ(5) etc. by the method of the sum of (1) gives the correct calculation result.
This can be considered because convergence is extremely slow in the case of complex numbers, but there is no tendency to converge at all.
Rather, it tends to diffuse.
In other words, it is inevitable to conclude that Riemann's hypothesis is a mistake.
We will fundamentally completely erroneous ones, For 150 years, We were trying to prove it.

**Category:** Number Theory

[1927] **viXra:1901.0101 [pdf]**
*submitted on 2019-01-09 00:16:39*

**Authors:** Johnny E. Magee

**Comments:** 7 Pages.

We identify equivalent restatements of the Brocard-Ramanujan diophantine equation, $(n! + 1) = m^2$; and employing the properties and implications of these equivalencies, prove that for all $n > 7$, there are no values of $n$ for which $(n! + 1)$ can be a perfect square.

**Category:** Number Theory

[1926] **viXra:1901.0046 [pdf]**
*submitted on 2019-01-04 11:35:20*

**Authors:** Nazihkhelifa

**Comments:** 4 Pages.

Relation between
The Euler Totient,
the counting prime formula
and the prime generating Functions
The theory of numbers is an area of mathematis hiih eals ith
the propertes of hole an ratonal numbers... In this paper I ill
intro uie relaton bet een Euler phi funiton an prime iountnn
an neneratnn formula, as ell as a ioniept of the possible
operatons e ian use ith them. There are four propositons hiih
are mentone in this paper an I have use the efnitons of these
arithmetial funitons an some Lemmas hiih refeit their
propertes, in or er to prove them

**Category:** Number Theory

[1925] **viXra:1901.0030 [pdf]**
*submitted on 2019-01-03 17:17:16*

**Authors:** Robert C. Hall

**Comments:** 24 Pages.

The Benford's Law Summation test consists of adding all numbers that begin with a particular first or first two digits and determining its distribution with respect to these first or first two digits numbers. Most people familiar with this test believe that the distribution is a uniform distribution for any distribution that conforms to Benford's law i.e. the distribution of the mantissas of the logarithm of the data set is uniform U[0,1). The summation test that results in a uniform distribution is true for an exponential function (geometric progression) but not true for a data set that conforms to a Log Normal distribution even when the Log Normal distribution itself closely approximates a Benford's Law distribution.

**Category:** Number Theory

[1924] **viXra:1901.0007 [pdf]**
*submitted on 2019-01-01 16:19:56*

**Authors:** Rachid Marsli

**Comments:** 11 Pages.

In this work, we show a sufficient and necessary condition for an integer of the form
(z^n-y^n)/(z-y)to be divisible by some perfect nth power p^n, where p is an odd prime. We also show how to construct such integers. A link between
the main result and Fermat’s last theorem is discussed. Other related ideas, examples and applications are provided.

**Category:** Number Theory

[1923] **viXra:1812.0497 [pdf]**
*submitted on 2018-12-31 12:45:35*

**Authors:** Pedro Hugo García Peláez

**Comments:** 5 Pages.

Series De Números Cuyos Factores Son La Lista De Los Números Primos Desde El Principio y Enumerando Todos Sin excepción

**Category:** Number Theory

[1922] **viXra:1812.0496 [pdf]**
*submitted on 2018-12-31 13:08:20*

**Authors:** Pedro Hugo García Peláez

**Comments:** 5 Pages.

Series of numbers whose factors are the list of prime numbers from the beginning and listing all without exception

**Category:** Number Theory

[1921] **viXra:1812.0495 [pdf]**
*submitted on 2018-12-31 14:29:57*

**Authors:** James Edwin Rock

**Comments:** Pages.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We show that in the interval (P_n, P_n^2) as P_n gets larger, there is an increasing number of twin primes.

**Category:** Number Theory

[1920] **viXra:1812.0494 [pdf]**
*submitted on 2018-12-31 09:51:52*

**Authors:** Simon Plouffe

**Comments:** 5 Pages.

We show here a new set of formulas for producing primes with a growth rate much smaller than the ones of Mills and Wright. Several examples of formulas are given. All results are empirical.

**Category:** Number Theory

[1919] **viXra:1812.0488 [pdf]**
*submitted on 2018-12-30 10:59:58*

**Authors:** Zeolla Gabriel Martín

**Comments:** 8 Pages.

This paper develops the analysis of Simple composite numbers by golden patterns. Examine how the Simple composite numbers are distributed in different combinations of multiples.

**Category:** Number Theory

[1918] **viXra:1812.0439 [pdf]**
*submitted on 2018-12-27 18:34:37*

**Authors:** Robert C. Hall

**Comments:** 5 Pages.

While it is fairly easy to prove that the Log Normal distribution becomes a Benford distribution as the standard deviation approaches infinity (see appendix A), it is a bit more difficult to prove that as the standard deviation approaches zero that the distribution becomes a Normal distribution with a mean of e^u where u is the mean of the natural logarithms of the data set values.

**Category:** Number Theory

[1917] **viXra:1812.0422 [pdf]**
*submitted on 2018-12-25 18:07:31*

**Authors:** A. A. Frempong

**Comments:** 9 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. The proof would be complete after showing that if A and B have a common prime factor, and C^z can be produced from the sum A^x + B^y. In the proof, one begins with A^x + B^y and change this sum to the single power, C^z as was done in the preliminaries. It was determined that if A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[1916] **viXra:1812.0418 [pdf]**
*submitted on 2018-12-24 06:01:30*

**Authors:** Nicolò Rigamonti

**Comments:** 15 Pages.

This paper shows the importance of two properties, which are at the base of the Riemann hypothesis. The key point of all the reasoning about the validity of the Riemann hypothesis is in the fact that only if the Riemann hypothesis is true, these two properties, which are satisfied by the non-trivial zeros, are both true. In fact, only if these two properties are both true , all non-trivial zeros lie on the critical line. Leave a comment, a reflection or an opinion about the paper. You shouldn’t keep your doubts and questions in yourself: if you think that it’s right or can’t attack it,share it, otherwise “ Why don’t you start a discussion?.........”.

**Category:** Number Theory

[1915] **viXra:1812.0400 [pdf]**
*submitted on 2018-12-24 05:12:29*

**Authors:** Jesús Sánchez

**Comments:** Pages.

b=0;
p=input('Input a number : ');
m=fix((p+1)/2);
for k=2:m+1;
fun=@(w) exp(p.*1j.*w).*exp(-2.*k.*1j.*w).*(1-exp(-m.*k.*1j.*w))./(1-exp(-k.*1j.*w));
a=integral(fun,-pi,pi);
b=b+a;
end;
b=b/(2*pi);
disp(b);
Above simple MATLAB/Octave program, can detect if a number is prime or not. If the result is zero (considering zero being less than 1e-5), the number introduced is a prime.
If the result is an integer, this result will tell us how many permutations of two divisors, the input number has. As you can check, no recurrent division by odd or prime numbers is done. Just this strange integral:
1/2π ∫_(-π)^π▒〖e^pjω (∑_(k=2)^(k=m+1)▒〖e^(-2kjω) ((1-e^(-mkjω))/(1-e^(-kjω) )) 〗)dω 〗
Being p the number that we want to check if it is a prime or not. And being m whatever integer number higher than (p+1)/2(the lowest, the most efficient the operation). As k and ω are independent variables, the sum can be taken outside the integral (as it is in the above program).
To get to this point, we will do the following. First, we will create a domain with all the composite numbers. This is easy, as you can just multiply one by one all the integers (greater or equal than 2) in that domain. So, you will get all the composite numbers (not getting any prime) in that domain.
Then, we will use the Fourier transform to change from this original domain (called discrete time domain in this regards) to the frequency domain. There, we can “ask”, using Parseval’s theorem, if a certain number is there or not. The use of Parseval’s theorem leads to the above integral. If the number p that we want to check is not in the domain, the result of the integral is zero and the number is a prime. If instead, the result is an integer, this integer will tell us how many permutations of two divisors the number p has. And, in consequence information how many factors, the number p has.
So, for any number p lower than 2m-1, you can check if it is prime or not, just making the numerical definite integration (but even this integral, if no further developments are done, the numerical integration is inefficient computing-wise compared with brute-force checking for example). To be added, is the question regarding the level of accuracy needed (number of decimals and number of steps in the numerical integration) to have a reliable result for large numbers. This will be commented on the paper, but a separate study will be needed to have detailed conclusions.
Of course, the best would be that in the future, an analytical result (or at least an approximation) for the summation or for the integration is achieved.

**Category:** Number Theory

[1914] **viXra:1812.0362 [pdf]**
*submitted on 2018-12-20 21:07:15*

**Authors:** Ho Soo Shin

**Comments:** 2 Pages. comments welcome!

As is well known, the Collatz sequence, which is also named as the hailstone sequence, follows the rule of Collatz conjecture. The rule requires us to divide any positive even integer by 2. We must multiply every positive odd number by 3 and then add 1 according to the rule. By investigating residues modulo 3, I will prove any integer multiple of 3 cannot appear more than one time in a Collatz sequence, which implies any multiple of 3 cannot be included in a possible cycle of the Collatz sequence.

**Category:** Number Theory

[1913] **viXra:1812.0340 [pdf]**
*submitted on 2018-12-19 17:50:09*

**Authors:** Wu ShengPing

**Comments:** 3 Pages.

The main idea of this article is simply calculating integer
functions in module. The algebraic in the integer modules is studied in
completely new style. By a careful construction the result that
two finite numbers is with unequal logarithms in a corresponding module is proven, which result is applied to solving
a kind of high degree diophantine equation.

**Category:** Number Theory

[1912] **viXra:1812.0322 [pdf]**
*submitted on 2018-12-18 09:16:24*

**Authors:** James Edwin Rock

**Comments:** 2 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License2

Let Pn be the n_th prime. For twin primes Pn – Pn-1 = 2. We give a heuristic argument that in the interval (Pn, Pn2) as Pn gets larger, there is an increasing number of twin primes.

**Category:** Number Theory

[1911] **viXra:1812.0312 [pdf]**
*submitted on 2018-12-19 05:32:58*

**Authors:** Julian Beauchamp

**Comments:** 3 Pages.

In 2002 Preda Mihailescu used the theory of cyclotomic fields and Galois modules to prove Catalan's Conjecture. In this short paper, we give a very simple proof. We first prove that no solutions exist for a^x-b^y=1 for a,b>1 and x,y>2. Then we prove that when x=2 the only solution for y is y=3.

**Category:** Number Theory

[1910] **viXra:1812.0300 [pdf]**
*submitted on 2018-12-17 23:40:30*

**Authors:** Aaron chau

**Comments:** 1 Page.

无论是在历史的任何时间，如不谈黎曼猜想则已；如谈，我们大家当然先要彻底弄明白：
黎曼临界线上从小到大的一组零点及其（零点空格），它们在数论上究竟是什么意思呢？

**Category:** Number Theory

[1909] **viXra:1812.0299 [pdf]**
*submitted on 2018-12-17 23:47:57*

**Authors:** Aaron chau

**Comments:** 2 Pages.

本文强调：质数与孪生质数分别都是无限的依据是算术中的（加减乘除）。
比如在古希腊，欧几里德证明质数无限，他所应用的是（乘除法）来表述反证法。
又比如，现时在伦敦来证明孪生质数无限，即无限存在二个质数的距离＝2；本文应用的是（加减法）来表述：就在（单数空格）里，单数与奇合数的个数分别在每一数段里的多与少。

**Category:** Number Theory

[1908] **viXra:1812.0296 [pdf]**
*submitted on 2018-12-18 01:33:20*

**Authors:** Joseph Olloh

**Comments:** 13 Pages. The content of the paper provides a fundamentally new way of looking at numbers.

We differentiate even and odd numbers into various groups and subgroups. We provide the properties of the forms of numbers which fall into each groups and subgroups. We expound on the relationship of a special group of even numbers and the collatz conjecture, we also derive an accurate formula to calculate the steps involved when an even number of the group is the initial value of the collatz operation.
For each group and subgroup of odd and even numbers, we discuss the observed pattern of their sequences and also derive accurate formulas for each sequence. Throughout, b, d, k, N, n, x, m, and z all denote positive integers, with d, and N denoting odd numbers, x and z denoting even numbers, and b denoting special even-even numbers
The order of priority of the properties of each group is key in the differentiation of the numbers into their various groups and subgroups.

**Category:** Number Theory

[1907] **viXra:1812.0287 [pdf]**
*submitted on 2018-12-16 09:20:54*

**Authors:** Zhang Tianshu

**Comments:** 14 Pages.

If regard positive integers which have a common prime factor as a kind, then the positive half line of the number axis consists of infinite many recurring segments which have same permutations of c kinds of integer’s points, where c≥1. In this article, the author shall prove Grimm’s conjecture by the method that changes orderly symbols of each kind of composite number’s points at the original number axis, so as to form consecutive composite number’s points within limits that proven Bertrand's postulate restricts.

**Category:** Number Theory

[1906] **viXra:1812.0242 [pdf]**
*submitted on 2018-12-13 08:45:16*

**Authors:** Jean Pierre Morvan

**Comments:** 6 Pages.

L'hypothèse de RIEMANN est fausse parce que la fonction zéta n'est pas nulle avec s = 1/2 +ib

**Category:** Number Theory

[1905] **viXra:1812.0235 [pdf]**
*submitted on 2018-12-13 23:53:47*

**Authors:** Toshiro Takami

**Comments:** 12 Pages.

As a prime number determination method, if the number is divided by 12 or 18 or 24 or 30 and the remainder is a prime number including 1, it is a prime number.
If the remainder is 2, it is an even number.
If the remainder is 3, it is a multiple of 3.
In particular, the remainder of the prime divided by 12 is limited to 1, 5,7 and 11.
Therefore, the method of dividing by 12 seems to be the most efficient.
There is also a method of using the fact that the remainder becomes a prime number of 30 or less by dividing by 30, which is better or more subtle.
When prime is divided by 30, the remainder always becomes prime.
Thus the prime number circulates in 12 or 30 and more for example, 18, 24, etc..
But, In the case of multiplication of prime number and prime number, the remainder may be 35 or 25 and prime number when dividing by 48.
And this is not only when dividing by 48, the same can be said when dividing by 30, 24, 18.
It can not be determined whether it is a prime or not, it is just a sieve.

**Category:** Number Theory

[1904] **viXra:1812.0208 [pdf]**
*submitted on 2018-12-11 16:27:17*

**Authors:** Kenneth A. Watanabe

**Comments:** 9 Pages.

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. The first mention of the Twin Prime Conjecture was in 1849, when de Polignac made the more general conjecture that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, I derive a function that corresponds to the number of twin primes less than n for large values of n. This equation is identical to that used to prove the Goldbach Conjecture. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using the same methodology, de Polignac’s conjecture is also shown to be true.

**Category:** Number Theory

[1903] **viXra:1812.0182 [pdf]**
*submitted on 2018-12-10 10:58:25*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 4 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we use the Fermat's Last Theorem (FLT) to give a proof of the ABC conjecture. We suppose that FLT is false ====> we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true ====> FLT is true. But, as FLT is true, then we deduce that the ABC conjecture is true.

**Category:** Number Theory

[1902] **viXra:1812.0154 [pdf]**
*submitted on 2018-12-08 16:12:41*

**Authors:** M. Sghiar

**Comments:** 7 Pages. french version © Copyright 2018 by M. Sghiar. All rights reserved. Respond to the author by email at: msghiar21@gmail.com

I show here that if $ x \in \mathbb{N}^*$ then $1 \in \mathcal{O}_S (x)= \{ S^n(x), n \in \mathbb{N}^* \} $ where $ \mathcal{O}_S (x)$ is the orbit of the function S defined on $\mathbb{R}^+$ by $S(x)= \frac{x}{2} + (x+\frac{1}{2}) sin^2(x\frac{\pi}{2})$, and I deduce the proof of the Syracuse conjecture.

**Category:** Number Theory

[1901] **viXra:1812.0150 [pdf]**
*submitted on 2018-12-08 21:32:29*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT simply for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not by way of contradiction) of FLT is based on our algebraic identity, denoted, {for convenience}, as $(r)^n+(s)^n=(t)^n$ with $r,s,t>0$ as functions of variables. We infer that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n+(t)^n\}=\{(x,y,z)|x,y,z\in\mathbb{N},(x)^n+y^n=z^n\}$ for $n\in\mathbb{N}, n>2$. In addition, we show, for integral values of $n>2$, that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n=t^n\}=\varnothing$. Hence, for $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[1900] **viXra:1812.0133 [pdf]**
*submitted on 2018-12-07 13:32:39*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License

Collatz sequences are formed by applying the Collatz algorithm to a positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. Eventually you get back to one. The Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[1899] **viXra:1812.0131 [pdf]**
*submitted on 2018-12-07 18:37:16*

**Authors:** Toshiro Takami

**Comments:** 5 Pages.

Calculationally, \sqrt(24a+1)=t is excellent as a prime number production formula.
But, \sqrt(24a+1)=t is a sequence that increases only by 2, 4, 2, 4 alternately.
The prime number may only be 2, 4, 2, 4 and the increasing sequence of complexity is increasing.
Occasionally, it may be prime that 2 and 4 combine to become 6, 8, 10, and so on.
\sqrt(12a+1)=t, \sqrt(8a+1)=t, \sqrt(6a+1)=t (a is positive integer)
as an examples.
\sqrt(24a+1)=t are exactly the same as these.
\sqrt(20a+1)=t, \sqrt(18a+1)=t, \sqrt(16a+1)=t, \sqrt(10a+1)=t, etc, tried, but thease did not become a prime number production formula at all.

**Category:** Number Theory

[1898] **viXra:1812.0130 [pdf]**
*submitted on 2018-12-07 21:04:48*

**Authors:** Zhang Tianshu

**Comments:** 12 Pages.

Since there are infinitely many consecutive satisfactory values of ε to enable A+B=C satisfying C>(rad(A, B, C))1+ε, thus the author uses a representative equality, namely 1+2N(2N-2)=(2N-1)2 satisfying (2N-1)2>[rad(1, 2N(2N-2), (2N-1)2)]1+ε, and that first let ε equal a value near the greater end of the infinitely many consecutive satisfactory values to prove the ABC conjecture; again let ε equal a value near the smaller end to negate the ABC conjecture. This shows that the ABC conjecture is in the ambiguity in which case of ε>0.

**Category:** Number Theory

[1897] **viXra:1812.0107 [pdf]**
*submitted on 2018-12-06 14:23:59*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 8 Pages. Comments welcome.

In this paper, we give the elliptic curve (E) given by the equation:
y^2=x^3+px+q
with $p,q \in Z$ not null simultaneous. We study a part of the conditions verified by $(p,q)$ so that it exists (x,y) \in Z^2 the coordinates of a point of the elliptic curve (E) given by the equation above.

**Category:** Number Theory

[1896] **viXra:1812.0074 [pdf]**
*submitted on 2018-12-04 16:03:11*

**Authors:** Stephen Marshall

**Comments:** 7 Pages.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1/2
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000).
It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann zeta function is defined for complex s with real part greater than 1 by the absolutely convergent infinite series:
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
The Riemann hypothesis asserts that all interesting solutions of the equation:
ζ(s) = 0
lie on a certain vertical straight line.
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn = 1 + 1/2+1/3+1/4+⋯+ 1/n = ∑_(n=1)^n▒1/n
Harmonic numbers have been studied since early times and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function.
The harmonic numbers roughly approximate the natural logarithm function and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers

**Category:** Number Theory

[1895] **viXra:1812.0071 [pdf]**
*submitted on 2018-12-04 22:22:02*

**Authors:** Colin James III

**Comments:** 2 Pages. © Copyright 2018 by Colin James III All rights reserved. Respond to the author by email at: info@ersatz-systems dot com.

Using the standard wiki definition of the Collatz conjecture, we map a positive number to imply that a divisor of two implies either an even numbered result (unchanged) or an odd numbered result (changed to the number multiplied by three plus one) to imply the final result of one. This is the shortest known confirmation of the conjecture, and in mathematical logic.

**Category:** Number Theory

[1894] **viXra:1812.0040 [pdf]**
*submitted on 2018-12-04 02:42:56*

**Authors:** Toshiro Takami

**Comments:** 3 Pages.

40.5+(n^2+1)/4 (n is positive integers)
This prime number production formula product integers and non-integers alternately, but picking up an integer portion yields a prime number up to the 40th number, and even after an integer that is not a prime number, many are prime numbers. Even in the region where prime numbers are prime, many are made prime.

**Category:** Number Theory

[1893] **viXra:1812.0020 [pdf]**
*submitted on 2018-12-01 15:11:29*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 6 Pages. Comments welcome.

In this paper, we assume that Beal conjecture is true, we give a complete proof of the ABC conjecture. We consider that Beal conjecture is false $\Longrightarrow$ we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true $\Longrightarrow$ Beal conjecture is true. But, if the Beal conjecture is true, then we deduce that the ABC conjecture is true

**Category:** Number Theory

[1892] **viXra:1812.0019 [pdf]**
*submitted on 2018-12-01 16:17:59*

**Authors:** Kamal Barghout

**Comments:** 10 Pages. This work is copyrighted material. No part of this work is to be copied without prior permission from the author

: A probabilistic proof of the Collatz conjecture is described via identifying a sequential permutation of even natural numbers by divisions by 2 that follows a recurrent pattern of the form x,1,x,1…, where x represents divisions by 2 more than once. The sequence presents a probability of 50:50 of divisions by 2 once as opposed to divisions by 2 more than once over the even natural numbers. The sequence also gives the same 50:50 probability of consecutive Collatz even elements when counted for division by 2 more than once as opposed to division by 2 once ratio of 2.8. Considering Collatz function producing random numbers, over sufficient iterations this probability distribution produces a descending order of its elements that leads to convergence of the Collatz function to the cycle 1-2-4-1.

**Category:** Number Theory

[1891] **viXra:1812.0018 [pdf]**
*submitted on 2018-12-01 17:00:45*

**Authors:** Khalid Ibrahim

**Comments:** 91 Pages.

In this paper, not only did we disprove the Riemann Hypothesis (RH) but also we showed that zeros of the Riemann zeta function $\zeta (s)$ can be found arbitrary close to the line $\Re (s) =1$. Our method to reach this conclusion is based on analyzing the fine behavior of the partial sum of the Dirichlet series with the Mobius function $M (s) = \sum_n \mu (n) /n^s$ defined over $p_r$ rough numbers (i.e. numbers that have only prime factors greater than or equal to $p_r$). Two methods to analyze the partial sum fine behavior are presented and compared. The first one is based on establishing a connection between the Dirichlet series with the Mobius function $M (s) $ and a functional representation of the zeta function $\zeta (s)$ in terms of its partial Euler product. Complex analysis methods (specifically, Fourier and Laplace transforms) were then used to analyze the fine behavior of partial sum of the Dirichlet series. The second method to estimate the fine behavior of partial sum was based on integration methods to add the different co-prime partial sum terms with prime numbers greater than or equal to $p_r$. Comparing the results of these two methods leads to a contradiction when we assume that $\zeta (s)$ has no zeros for $\Re (s) > c$ and $c <1$.

**Category:** Number Theory

[1890] **viXra:1812.0002 [pdf]**
*submitted on 2018-12-01 05:01:55*

**Authors:** Zhang Tianshu

**Comments:** 21 Pages.

In this article, the author first classify A, B and C according to their respective odevity, and thereby get rid of two kinds which belong not to AX+BY=CZ. Then, affirm the existence of AX+BY=CZ in which case A, B and C have at least a common prime factor by several concrete equalities. After that, prove AX+BY≠CZ in which case A, B and C have not any common prime factor by the mathematical induction with the aid of the distinct odd-even relation on the premise whereby even number 2W-1HZ as symmetric center of positive odd numbers concerned after divide the inequality in four. Finally, reach a conclusion that the Beal’s conjecture holds water via the comparison between AX+BY=CZ and AX+BY≠CZ under the given requirements.

**Category:** Number Theory

[1889] **viXra:1811.0503 [pdf]**
*submitted on 2018-11-29 17:48:12*

**Authors:** Zhang Tianshu

**Comments:** 18 Pages.

In this article, the author uses the mathematical induction, classifies positive integers gradually, and passes necessary operations by the operational rule to achieve finally the purpose proving Collatz conjecture.

**Category:** Number Theory

[1888] **viXra:1811.0476 [pdf]**
*submitted on 2018-11-28 19:00:23*

**Authors:** 9

**Comments:** 9 Pages.

I found out
(√10a×1-1) (√10a×1+1)+20
(a is positive integer.)
is very effective of prime number production equation.

**Category:** Number Theory

[1887] **viXra:1811.0471 [pdf]**
*submitted on 2018-11-29 00:56:11*

**Authors:** Divyendu Priyadarshi

**Comments:** 1 Page.

In this small paper , I have given a simple proof of
already well established fact that there are infinitely many prime numbers.

**Category:** Number Theory

[1886] **viXra:1811.0459 [pdf]**
*submitted on 2018-11-27 10:05:16*

**Authors:** Victor Sorokine

**Comments:** 2 Pages.

he essence of the proof of the FLT:
The first case (ABC is not a multiple of n):
In one of the equivalent Fermat equations, the 3rd digit of the sum of the powers of the last digits of the bases greater than 1, which cannot be zeroed using the second digits with the sum of the latter equal to 0 or n-1.
+++
The second case (A or B or C is a multiple of n):
(k+2)-th digit in the number D=(A+B)^n-(C-B)^n-(C-A)^n, where the number A+B-C ends by k zeros, is not zero, but after adding to the number D zero as 0=A^n+B^n-C^n (k+2)-th digit is zero.

**Category:** Number Theory

[1885] **viXra:1811.0457 [pdf]**
*submitted on 2018-11-27 10:06:38*

**Authors:** Victor Sorokine

**Comments:** 2 Pages. Russian version

Суть доказательства ВТФ:
Первый случай (ABC не кратно n):
В одном из эквивалентных равенств Ферма 3-я цифра суммы степеней последних цифр оснований больше 1, которую невозможно обнулить с помощью вторых цифр с суммой последних, равной 0 или n-1.
+++
Второй случай (A или B или C кратно n):
(k+2)-я цифра в числе D=(A+B)^n-(C-B)^n-(C-A)^n, где число A+B-C оканчивается на k нулей, не равна нулю, но после прибавления к числу D нуля в виде 0=A^n+B^n-C^n (k+2)-я цифра равна нулю.

**Category:** Number Theory

[1884] **viXra:1811.0348 [pdf]**
*submitted on 2018-11-23 04:44:14*

**Authors:** Toshiro Takami

**Comments:** 10 Pages.

Last time, I expressed ζ (odd number), such as ζ(3), ζ (5), ζ(7), ζ(9), ζ(11), ζ(13), ζ (15), ζ(17), ζ (19), ζ (21), ζ(23) and made an official.
Another way of expressing ζ (odd number), such as ζ(3), ζ (5), ζ(7) ζ(9), ζ(11), ζ(13), ζ(15), ζ(17), ζ(19), ζ(21), ζ(23) and made an official.

**Category:** Number Theory

[1883] **viXra:1811.0338 [pdf]**
*submitted on 2018-11-21 16:41:59*

**Authors:** Marco Ripà

**Comments:** 3 Pages.

We provide a preliminary proof of Ripà’s Conjecture 3 about the convergence speed of tetration, published in October 2018, which states that, for any natural number "v", exists (at least) another natural number "a", not a multiple of 10, such that V(a)=v, where V(a) represents the convergence speed of the tetration a^^b.

**Category:** Number Theory

[1882] **viXra:1811.0320 [pdf]**
*submitted on 2018-11-20 09:10:58*

**Authors:** Zeolla Gabriel Martín

**Comments:** 9 Pages. English language

This paper develops a new multiplication algorithm that works absolutely with all the numbers.

**Category:** Number Theory

[1881] **viXra:1811.0291 [pdf]**
*submitted on 2018-11-18 05:43:55*

**Authors:** Toshiro Takami

**Comments:** 7 Pages.

Prime number equation
a=\frac{t^2+232}{8}
a=\frac{t^2+93}{6}
(t is positive integer)
thease contain not prime numbers, but many are prime numbers.
And thase are contain number with decimal point, I pulled out only integers.
Write as halfway progress.

**Category:** Number Theory

[1880] **viXra:1811.0288 [pdf]**
*submitted on 2018-11-18 11:45:11*

**Authors:** Robert C. Hall

**Comments:** 20 Pages.

The summation test consists of adding all numbers that begin with a particular first digit or first two digits and determining its distribution with respect to these first or first two digits numbers. Most people familiar with this test believe that the distribution is a uniform distribution for any distribution that conforms to Benford's law i.e. the distribution of the mantissas of the logarithm of the data set is uniform U[0,1). The summation test that results in a uniform distribution is true for an exponential function (geometric progression) i.e. y = a exp(kt) but not necessarily true for other data sets that conform exactly to Benford'a law.

**Category:** Number Theory

[1879] **viXra:1811.0282 [pdf]**
*submitted on 2018-11-18 21:44:13*

**Authors:** Stefan Bereza

**Comments:** 4 Pages.

Fermat's Last Theorem (FLT) x^p + y^p = z^p could be seen as a special case of more generalized Beal's Conjecture (BC) x^m + y^n = z^r. Those equations are impossible when x, y and z are natural numbers and coprimes and {p, m, n, r}> = 3; if m = n = r (= p), then it is FLT; if not, Beal's Conjecture.
In BC, if x, y and z are integers and have a common factor, they can be measured (without rest) with this factor as a common unit - making x, y and z in the equation rational to each other. FLT can be proved with proving irrationality of triangles inscribed into an ellipse whose sides x, y and z represent the Fermat's equation x^p + y^p = z^p ; here, for x, y and z a common unit cannot be found. The BC equation
x^m + y^n = z^r (without a common factor) can be simplified to the Fermat's equation x^p + y^p = z^p which - at the lacking common unit - makes x, y and z impossible to be all rational to each other.

**Category:** Number Theory

[1878] **viXra:1811.0263 [pdf]**
*submitted on 2018-11-17 20:39:55*

**Authors:** Olivier Massot

**Comments:** 22 Pages.

The binomial formula, set by Isaac Newton, is of utmost importance and has been extensively used in many different fields. This study (in French) aims at coming up with alternative expressions to the Newton's formula.

**Category:** Number Theory

[1877] **viXra:1811.0262 [pdf]**
*submitted on 2018-11-17 20:44:21*

**Authors:** Olivier Massot

**Comments:** 22 Pages.

The binomial formula, set by Isaac Newton, is of utmost importance and has been extensively used in many different fields. This study aims at coming up
with alternative expressions to the Newton's formula.

**Category:** Number Theory

[1876] **viXra:1811.0250 [pdf]**
*submitted on 2018-11-16 11:08:43*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 6 Pages. Comments welcome.

In this paper, we give a proof minus $\epsilon$ of the $ABC$ conjecture, considering that Beal conjecture is true. Some conditions are proposed for the proof, perhaps it needs some justifications that is why I give the title of the paper " a proof minus $\epsilon$ of the $ABC$ conjecture".

**Category:** Number Theory

[1875] **viXra:1811.0211 [pdf]**
*submitted on 2018-11-13 08:01:46*

**Authors:** Zeolla Gabriel Martín

**Comments:** 9 Pages. Idioma: Español

Este documento desarrolla y demuestra el descubrimiento de un nuevo algoritmo de multiplicación que funciona absolutamente con todos los números.

**Category:** Number Theory

[1874] **viXra:1811.0179 [pdf]**
*submitted on 2018-11-11 19:03:33*

[1873] **viXra:1811.0159 [pdf]**
*submitted on 2018-11-11 04:13:06*

**Authors:** Zach Don

**Comments:** 1 Page.

In this paper, I will be presenting an alternative way of writing the Riemann zeta function in terms of Euler's constant, e.

**Category:** Number Theory

[1872] **viXra:1811.0158 [pdf]**
*submitted on 2018-11-11 04:19:06*

**Authors:** Zach Don

**Comments:** 1 Page.

In this paper, I will propose a legitimate way of re-writing the Riemann zeta function in terms of Euler's constant, e.

**Category:** Number Theory

[1871] **viXra:1811.0119 [pdf]**
*submitted on 2018-11-07 10:57:26*

**Authors:** Viktor Strohm

**Comments:** 5 Pages.

In accordance with the General Theory Systems of Urmantsev (GTSU) [1, 2, 3], the set of primes is considered as a system of objects. For the relationship between objects taken the difference of prime numbers. Revealed periodicity of pairs of intervals.

**Category:** Number Theory

[1870] **viXra:1811.0116 [pdf]**
*submitted on 2018-11-07 11:25:55*

**Authors:** Colin James III

**Comments:** 1 Page. © Copyright 2018 by Colin James III All rights reserved. Respond to the author by email at: info@ersatz-systems dot com.

Properties of the zeta function of the Riemann hypothesis are not confirmed as tautologous and hence refute it.

**Category:** Number Theory

[1869] **viXra:1811.0112 [pdf]**
*submitted on 2018-11-07 18:02:13*

**Authors:** Es-said En-naoui

**Comments:** 5 Pages.

The Goldbach conjecture dates back to 1742 ; we refer the reader to [1]-[2] for a history of the conjecture. Christian Goldbach stated that every odd integer greater than seven can be written as the sum of at most three prime numbers. Leonhard Euler then made a stronger conjecture that every even integer greater than four can be written as the sum of two primes. Since then, no one has been able to prove the Strong Goldbach Conjecture.\\
The only best known result so far is that of Chen [3], proving that every sufficiently large even integer N can be written as the sum of a prime number and the product of at most two prime numbers. Additionally, the conjecture has been verified to be true for all even integers up to $4.10^{18}$ in 2014 , J\"erg [4] and Tom\'as [5]. In this paper, we prove that the conjecture is true for all even integers greater than 8.

**Category:** Number Theory

[1868] **viXra:1811.0080 [pdf]**
*submitted on 2018-11-05 12:05:54*

**Authors:** Ralf Wüsthofen

**Comments:** 1 Page.

This note proves the inconsistency of the Peano arithmetic (PA) by deriving both the strong Goldbach conjecture and its negation.

**Category:** Number Theory

[1867] **viXra:1811.0072 [pdf]**
*submitted on 2018-11-06 02:39:26*

**Authors:** Quang Nguyen Van

**Comments:** 4 Pages.

We give expression of w^n and the possible to apply for solving Fermat's Last theorem.

**Category:** Number Theory

[1866] **viXra:1811.0046 [pdf]**
*submitted on 2018-11-03 21:04:42*

**Authors:** Colin James III

**Comments:** 1 Page. © Copyright 2018 by Colin James III All rights reserved. Respond to the author by email at: info@ersatz-systems dot com.

Karush-Kuhn-Tucker constraints for linear programming are not tautologous.

**Category:** Number Theory

[1865] **viXra:1811.0032 [pdf]**
*submitted on 2018-11-02 17:55:05*

**Authors:** Hassine Saidane

**Comments:** 4 Pages.

Optimization of relevant concepts such as action or utility functions enabled the derivation of theories and laws in some scientific fields such as physics and economics. This fact suggested that the problem of the location of the Riemann Zeta Function’s (RZF) nontrivial zeros can be addressed in a mathematical programming framework. Using a constrained nonlinear optimization formulation of the problem, we prove that RZF’s nontrivial zeros are located on the critical line, thereby confirming the Riemann Hypothesis. This result is a direct implication of the Kuhn-Tucker necessary optimality conditions for the formulated nonlinear program.
Keywords: Riemann Zeta function, Riemann Hypothesis, Optimization, Kuhn-Tucker conditions

**Category:** Number Theory

[1864] **viXra:1811.0031 [pdf]**
*submitted on 2018-11-02 18:27:16*

**Authors:** César Aguilera

**Comments:** 9 pages, 4 figures, 4 tables.

A set of relations between perfect numbers is presented. Then some properties of this relations and how they behave, next, a geometric interpretation, a function and finally, the way this function works.

**Category:** Number Theory

[1863] **viXra:1811.0026 [pdf]**
*submitted on 2018-11-03 05:56:56*

**Authors:** Toshiro Takami

**Comments:** 25 Pages.

All prime number expressed
18a + p (a is integer include 0, p is prime number less than 18 include 1, and a continues to infinity).
Prime numbers cycle at 18.
24a + p (a is integer include 0, p is prime number less than 24 include 1, and a continues to infinity).
Prime numbers cycle at 24.
and 30a + p (a is integer include 0, p is prime number less than 30 include 1, and a continues to infinity).
Prime numbers cycle at 30.
There is no exception.
I noticed while building prime number production formula. Prime numbers arise from several lines. Therefore, I believe it is extremely difficult to build a prime number production formula.

**Category:** Number Theory

[1862] **viXra:1811.0017 [pdf]**
*submitted on 2018-11-01 12:03:50*

**Authors:** Salvatore Gerard Micheal

**Comments:** 2 Pages.

a brief exposition on quantifying irrational density within the reals - and - attempt to categorize groups of irrationals

**Category:** Number Theory

[1861] **viXra:1811.0016 [pdf]**
*submitted on 2018-11-01 12:17:02*

**Authors:** Clemens Kroll

**Comments:** 6 Pages.

It is shown that Riemann’s hypothesis is true by showing that an equivalent statement is true. Even more, it is shown that Stieltjes’ conjecture is true.

**Category:** Number Theory

[977] **viXra:1902.0188 [pdf]**
*replaced on 2019-02-18 21:44:57*

**Authors:** Toshiro Takami

**Comments:** 4 Pages.

I proved Riemann hypothesis.
It proved that it never takes a zero point if a<0.5, 0.5<a.
There are many zeros such as 0.5+i14.1347, but all the known zero points are on the 0.5 axis.
I used the smallest prime number 2.

**Category:** Number Theory

[976] **viXra:1902.0188 [pdf]**
*replaced on 2019-02-15 20:32:47*

**Authors:** Toshiro Takami

**Comments:** 9 Pages.

I proved the Riemann hypothesis.
0.5 is the vertical groove, the imaginary value of the nontrivial zero is the horizontal groove, the place where this intersects is a non-trivial zero point.
Therefore, the nontrivial zero point always exists on 0.5.
When z = a + ib, z takes the minimum value when a value is 0.5, and when real value of b is non-trivial zero value, real value and imaginary value take the minimum value.
Also, z takes the minimum real value and imaginary value when z is 0.5 when the value of b is a value of a nontrivial zero.

**Category:** Number Theory

[975] **viXra:1902.0117 [pdf]**
*replaced on 2019-02-16 16:23:30*

**Authors:** Toshiro Takami

**Comments:** 10 Pages.

Although the calculation result of zeta comes out immediately, it was found out that it seems that it calculates by the expression as shown below when examining what kind of calculation method is used integrally.
If I had calculated according to the method in Introduction, I thought that it would take a very long time to calculate zeta, I studied mysteriously.
However, it is judgmental as to whether this investigation result is true or not.
If this calculation method is used, it is natural that a zero point appears only on 0.5.
As an overview of the net, it seems that the following formula is frequently used for calculation of Riemann zeta function.
In this calculation method, there is no zero point other than 0.5.
The mystery of 150 years has been solved.
End of Riemann Hypothesis proof.
ζ(s)＝π^(s-1/2)Γ((1-s)/2)／Γ(s/2)ζ(1-s), s=a+bi

**Category:** Number Theory

[974] **viXra:1902.0117 [pdf]**
*replaced on 2019-02-14 18:03:45*

**Authors:** Toshiro Takami

**Comments:** 3 Pages.

There is no infinity. The world of numbers is a sphere, not a two dimensional world represented by paper.
Therefore, 0 is present beyond eternity.
Running the Earth eternally forever will come back to the original place. The number is made like this.
Infinite next is 0.
Equal Infinite next is x=y=imaginary number=0.
Therefore Riemann Hypothesis is denied.

**Category:** Number Theory

[973] **viXra:1902.0117 [pdf]**
*replaced on 2019-02-12 01:24:03*

**Authors:** Toshiro Takami

**Comments:** 24 Pages.

2^s/(2^-1)*3^s/(3^-1)*5^s/(5^s-1)*7^s/(7^s-1)………
Whether the above equation converges to 0 was verified.
Convergence is extremely slow, and divergence tendency was rather rather abundant when the prime number was 1000 or more.
It was thought that the above equation could possibly be an expression that can be composed only of real numbers.

**Category:** Number Theory

[972] **viXra:1902.0117 [pdf]**
*replaced on 2019-02-07 06:22:01*

**Authors:** Toshiro Takami

**Comments:** 3 Pages.

repeat zeta, the imaginary value goes to 0.
This is considered to be one piece of evidence that the analytic connection is a monster.
We must get out of the brainwashing of analytic connection monsters as soon as possible.

**Category:** Number Theory

[971] **viXra:1902.0106 [pdf]**
*replaced on 2019-02-10 10:53:42*

**Authors:** Algirdas Antano Maknickas

**Comments:** 2 Pages.

This remark gives analytical solution of Last Fermat's Theorem

**Category:** Number Theory

[970] **viXra:1902.0106 [pdf]**
*replaced on 2019-02-09 02:11:25*

**Authors:** Algirdas Antano Maknickas

**Comments:** 2 Pages. In previous version was mistake in middle name.

This remark gives analytical solution of Last Fermat's Theorem

**Category:** Number Theory

[969] **viXra:1902.0020 [pdf]**
*replaced on 2019-02-03 20:29:25*

**Authors:** Toshiro Takami

**Comments:** 174 Pages.

I also found a zero point which seems to deviate from 0.5.
I thought that the zero point outside 0.5 can not be found very easily in the area which can not be shown in the figure, but this area can not be represented in the figure but can be found one after another.
It is completely unknown whether this axis is distorted in the 0.5 axis or just by coincidence.
The number of zero points in the area that can not be shown in the figure is now 43.
No matter how you looked it was not found in other areas.
It seemed that there is no other way to interpret this axis as 0.5 axis is distorted in this area.
Somewhere on the net there is a memory that reads the mathematician's view that "there are countless zero points in the vicinity of 0.5 on high area".
We are reporting that the zero point search of the high-value area of the imaginary part which was giving up as it is no longer possible with the supercomputer is no longer possible, is reported.
43 zero-point searches in the high-value area of the imaginary part are thus successful.
This means that the zero point search in the high-value area of the imaginary part has succeeded in the 43.
We will also write 43 zero point searches of the successful high-value area of the imaginary part.
There are many counterexamples far beyond 0.5, which is far beyond the limit, but the computer can not calculate it.
Moreover, I believe that it can only be confirmed on supercomputer whether this is really counterexample. In addition, it is necessary to make corrections in the supercomputer.

**Category:** Number Theory

[968] **viXra:1902.0005 [pdf]**
*replaced on 2019-02-19 08:32:39*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 7 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let P_n be the n_th prime. For twin primes P_n – P_(n-1) = 2. Let X be the number of (6j –1, 6j+1) pairs in the interval [P_n, P_n^2]. The number of twin primes (TPAn) in [P_n, P_n^2] can be approximated by the formula
(a_3 /5)(a_4 /7)(a_5 /11)…(a_n /P_n)(X) for 3 ≤ m ≤ n, a_m = P_m –2 .
We establish a lower bound for TPAn (3/5)(5/7)(7/9)…(P_n–2)/P_n)(X) = 3X/P_n < TPAn.
We exhibit a formula showing as P_n increases, the number of twin primes in the interval [P_n, P_n^2] also increases. Let P_n – P_(n-1) = c. For all n (TPAn-1)(1+(2c –2)/2P_(n-1)+(c^2–2c)/2P_(n-1)^2) < TPAn

**Category:** Number Theory

[967] **viXra:1902.0005 [pdf]**
*replaced on 2019-02-16 12:01:31*

**Authors:** James Edwin Rock

**Comments:** 2 Pages of exposition and 7 pages of graphs

Let P_n be the n_th prime. For twin primes P_n – P_(n-1) = 2. Let X be the number of (6j –1, 6j+1) pairs in the interval [P_n, P_n^2]. The number of twin primes (TPAn) in [P_n, P_n^2] can be approximated by the formula
(a_3 /5)(a_4 /7)(a_5 /11)…(a_n /P_n)(X) for 3 ≤ m ≤ n, a_m = P_m –2 .
We establish a lower bound for TPAn (3/5)(5/7)(7/9)…(P_n–2)/P_n)(X) = 3X/P_n < TPAn.
We exhibit a formula showing as P_n increases, the number of twin primes in the interval [P_n, P_n^2] also increases. Let P_n – P_(n-1) = c. For all n (TPAn-1)(1+(2c –2)/2P_(n-1)+(c^2–2c)/2P_(n-1)^2) < TPAn

**Category:** Number Theory

[966] **viXra:1902.0005 [pdf]**
*replaced on 2019-02-13 08:29:18*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 6 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

**Category:** Number Theory

[965] **viXra:1901.0436 [pdf]**
*replaced on 2019-02-13 08:35:02*

**Authors:** Kenneth A. Watanabe

**Comments:** 8 Pages.

Legendre's conjecture, states that there is a prime number between n^2 and (n + 1)^2 for every positive integer n. In this paper, an equation was derived that determines the number of prime numbers less than n for large values of n. Then it is proven by mathematical induction that there is at least one prime number between n^2 and (n + 1)^2 for all positive integers n thus proving Legendre’s conjecture.

**Category:** Number Theory

[964] **viXra:1901.0436 [pdf]**
*replaced on 2019-02-11 08:33:18*

**Authors:** Kenneth A. Watanabe

**Comments:** 7 Pages.

Legendre's conjecture, states that there is a prime number between n^$ and (n + 1)^2 for every positive integer n. In this paper, an equation was derived that determines the number of prime numbers less than n for large values of n. Then it is proven by mathematical induction that there is at least 1 prime number between n^2 and (n + 1)^2 for all positive integers n thus proving Legendre’s conjecture.

**Category:** Number Theory

[963] **viXra:1901.0430 [pdf]**
*replaced on 2019-01-31 12:44:22*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Paper corrected of a mistake in the last version. Comments welcome.

In this paper, we consider the abc conjecture, then we give a proof of the conjecture c<rad^2(abc) that it will be the key to the proof of the abc conjecture.

**Category:** Number Theory

[962] **viXra:1901.0430 [pdf]**
*replaced on 2019-01-29 08:20:57*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we consider the ABC conjecture, then we give a proof that C<rad*2(ABC) that it will be the key of the proof of the ABC conjecture.

**Category:** Number Theory

[961] **viXra:1901.0227 [pdf]**
*replaced on 2019-02-21 08:01:40*

**Authors:** James Edwin Rock

**Comments:** 12 pages. Section 5 is rewritten.This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Collatz sequences are formed by applying the Collatz algorithm to any positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. If the Collatz conjecture is true eventually you always get back to one. A connected Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[960] **viXra:1901.0227 [pdf]**
*replaced on 2019-01-29 11:21:03*

**Authors:** James Edwin Rock

**Comments:** 7 pages. This paper should be much easier to understand than the paper it is replacing

Collatz sequences are formed by applying the Collatz algorithm to any positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. If the Collatz conjecture is true eventually you always get back to one. A connected Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[959] **viXra:1901.0227 [pdf]**
*replaced on 2019-01-17 08:32:12*

**Authors:** James Edwin Rock

**Comments:** 5 pages. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

**Category:** Number Theory

[958] **viXra:1901.0104 [pdf]**
*replaced on 2019-02-15 21:57:44*

**Authors:** Toshiro Takami

**Comments:** 8 Pages.

I tried to prove that(Riemann hypothesis), but I realized that I can not prove how I did it.
When we calculate by the sum method of (1) we found that the nontrivial zero point will never converge to zero.
Calculating ζ(2), ζ(3), ζ(4), ζ(5) etc. by the method of the sum of (1) gives the correct calculation result.
This can be considered because convergence is extremely slow in the case of complex numbers, but there is no tendency to converge at all. Rather, it tends to diffuse.
In other words, it is inevitable to conclude that Riemann's hypothesis is a mistake.
We will fundamentally completely erroneous ones, For 150 years, We were trying to prove it.

**Category:** Number Theory

[957] **viXra:1901.0104 [pdf]**
*replaced on 2019-02-06 13:56:01*

**Authors:** Toshiro Takami

**Comments:** 3 Pages.

In the mathematical world there was a crawl of analytical connections walking around and I noticed the fact that the mathematics world was dominated by the monster.
You should wake up quickly escape from the brainwashing analysis connection (monster).
I notice this, I will dare to write for justice.

**Category:** Number Theory

[956] **viXra:1901.0104 [pdf]**
*replaced on 2019-02-04 00:52:16*

**Authors:** Toshiro Takami

**Comments:** 184 Pages.

I also found a zero point which seems to deviate from 0.5.
I thought that the zero point outside 0.5 can not be found very easily in the area which can not be shown in the figure, but this area can not be represented in the figure but can be found one after another.
It is completely unknown whether this axis is distorted in the 0.5 axis or just by coincidence.
The number of zero points in the area that can not be shown in the figure is now 43.
No matter how you looked it was not found in other areas.
It seemed that there is no other way to interpret this axis as 0.5 axis is distorted in this area.
Somewhere on the net there is a memory that reads the mathematician's view that "there are countless zero points in the vicinity of 0.5 on high area".
We are reporting that the zero point search of the high-value area of the imaginary part which was giving up as it is no longer possible with the supercomputer is no longer possible, is reported.
43 zero-point searches in the high-value area of the imaginary part are thus successful.
This means that the zero point search in the high-value area of the imaginary part has succeeded in the 43.
We will also write 43 zero point searches of the successful high-value area of the imaginary part.
There are many counterexamples far beyond 0.5, which is far beyond the limit, but the computer can not calculate it.
Moreover, I believe that it can only be confirmed on supercomputer whether this is really counterexample. In addition, it is necessary to make corrections in the supercomputer.

**Category:** Number Theory

[955] **viXra:1901.0104 [pdf]**
*replaced on 2019-02-02 16:39:04*

**Authors:** Toshiro Takami

**Comments:** 182 Pages.

I also found a zero point which seems to deviate from 0.5.
I thought that the zero point outside 0.5 can not be found very easily in the area which can not be shown in the figure, but this area can not be represented in the figure but can be found one after another.
It is completely unknown whether this axis is distorted in the 0.5 axis or just by coincidence.
The number of zero points in the area that can not be shown in the figure is now 39.
The following are the 39 points.

**Category:** Number Theory

[954] **viXra:1901.0104 [pdf]**
*replaced on 2019-01-29 23:35:22*

**Authors:** Toshiro Takami

**Comments:** 4 Pages.

I tried to prove that(Riemann hypothesis), but I realized that I can not prove how I did it.
When we calculate by the sum method of (1) we found that the nontrivial zero point will never converge to zero.
Calculating ζ(2), ζ(3), ζ(4), ζ(5) etc. by the method of the sum of (1) gives the correct calculation result.
This can be considered because convergence is extremely slow in the case of complex numbers, but there is no tendency to converge at all. Rather, it tends to diffuse.
In other words, it is inevitable to conclude that Riemann's hypothesis is a mistake.
We will fundamentally completely erroneous ones, For 150 years, We were trying to prove it.

**Category:** Number Theory

[953] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-30 16:38:19*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. This replacement paper adds a ratio TPA / 2.06 column to Table 2 and Table 1 contains a comparison of the twin prime pairs formula for [743,743^2] and [19993,19993^2]. It gives a better explanation for the twin prime pairs formula.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We exhibit formula for calculating the number of twin primes in the closed interval [P_n, P_n^2]. We prove that as P_n increases the number of twin primes in the interval [P_n, Pn^2] also increases.

**Category:** Number Theory

[952] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-25 15:03:02*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 3 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We exhibit formula for calculating the number of twin primes in the closed interval [P_n, P_n^2]. We prove that as P_n increases the number of twin primes in the interval [P_n, Pn^2] also increases.

**Category:** Number Theory

[951] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-22 09:05:33*

**Authors:** James Edwin Rock

**Comments:** 5 Pages.

Let Pn be the n_th prime. For twin primes Pn – Pn-1 = 2. We exhibit two formulas for calculating the number of twin primes in the closed interval [Pn, Pn^2]. We show there is a lower limit for the number of twin primes in the closed interval [Pn, Pn^2].

**Category:** Number Theory

[950] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-14 10:52:26*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 3 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let Pn be the n_th prime. For twin primes Pn – Pn-1 = 2. We exhibit two formulas for calculating the number of twin primes in the closed interval [Pn, Pn^2]. We show there is a lower limit for the number of twin primes in the closed interval [Pn, Pn^2].

**Category:** Number Theory

[949] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-07 12:49:24*

**Authors:** James Edwin Rock

**Comments:** 1 page of exposition and 3 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We exhibit two formulas for calculating the number of twin primes in the closed interval [P_n, P_n^2]. We show there is a lower limit for the number of twin primes in the closed interval [P_n, Pn^2].

**Category:** Number Theory

[948] **viXra:1812.0494 [pdf]**
*replaced on 2019-01-18 17:04:43*

**Authors:** Simon Plouffe

**Comments:** 10 Pages.

A set of new formulas for primes are given. These formulas have a growth rate much smaller than the ones of Mills and Wright.

**Category:** Number Theory

[947] **viXra:1812.0494 [pdf]**
*replaced on 2019-01-07 23:07:56*

**Authors:** Simon Plouffe

**Comments:** 7 Pages.

In 1947, W. H. Mills published a paper describing a formula that gives primes : if A = 1.3063778838630806904686144926… then ⌊A^(3^n ) ⌋ is always prime, here ⌊x⌋ is the integral part of x. Later in 1951, E. M. Wright published another formula, if g_0=α = 1.9287800… and g_(n+1)=2^(g_n ) then
⌊g_n ⌋= ⌊2^(…2^(2^α ) ) ⌋ is always prime.
The primes are uniquely determined by α , the prime sequence is 3, 13, 16381, …
The growth rate of these functions is very high since the fourth term of Wright formula is a 4932 digit prime and the 8’th prime of Mills formula is a 762 digit prime.
A new set of formulas is presented here, giving an arbitrary number of primes minimizing the growth rate. The first one is : if a_0=43.8046877158…and a_(n+1)= 〖〖〖(a〗_n〗^(5/4))〗^n , then if S(n) is the rounded values of a_n, S(n) = 113,367,102217,1827697,67201679,6084503671, …. Other exponents can also give primes like 11/10, or 101/100. If a_0 is well chosen then it is conjectured that any exponent > 1 can also give an arbitrary series of primes. The method for obtaining the formulas is explained. All results are empirical.

**Category:** Number Theory

[946] **viXra:1812.0494 [pdf]**
*replaced on 2018-12-31 12:31:57*

**Authors:** Simon Plouffe

**Comments:** 6 Pages.

A new set of formulas for primes is presented, they are better than the ones of Mills and Wright.

**Category:** Number Theory

[945] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-15 03:19:29*

**Authors:** A. A. Frempong

**Comments:** 11 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. In the numerical equations, two approaches have been used to change the sum, A^x + B^y, of two powers to a single power, C^z. In one approach, the application of factorization is the main principle, while in the other approach, a formula derived from A^x + B^y was applied. The two approaches changed the sum A^x + B^y to a single power, C^z, perfectly. The derived formula confirmed the validity of the assumption that it is necessary that the sum A^x + B^y has a common prime factor before C^z can be derived. It was concluded that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor.

**Category:** Number Theory

[944] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-08 13:30:23*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[943] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-05 16:32:17*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[942] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-04 11:56:04*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[941] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-01 21:14:39*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[940] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-01 03:48:04*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. Using concrete examples, the proof would be complete after showing that if A and B have a common prime factor, the sum A^x + B^y can be changed to a product and to a single power, C^z. In the proof, one begins with A^x + B^y and changes this sum to the single power, C^z . It was determined that if A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[939] **viXra:1812.0422 [pdf]**
*replaced on 2018-12-28 14:18:23*

**Authors:** A. A. Frempong

**Comments:** 8 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. The proof would be complete after showing that if A and B have a common prime factor, and C^z can be produced from the sum A^x + B^y. In the proof, one begins with A^x + B^y and changes this sum to the single power, C^z as was done in the preliminaries. It was determined that if A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[938] **viXra:1812.0400 [pdf]**
*replaced on 2018-12-25 07:45:30*

**Authors:** Jesús Sánchez

**Comments:** 21 Pages.

b=0;
p=input('Input a number : ');
m=fix((p+1)/2);
for k=2:m+1;
fun=@(w) exp(p.*1j.*w).*exp(-2.*k.*1j.*w).*(1-exp(-m.*k.*1j.*w))./(1-exp(-k.*1j.*w));
a=integral(fun,-pi,pi);
b=b+a;
end;
b=b/(2*pi);
disp(b);
Above simple MATLAB/Octave program, can detect if a number is prime or not. If the result is zero (considering zero being less than 1e-5), the number introduced is a prime.
If the result is an integer, this result will tell us how many permutations of two divisors, the input number has. As you can check, no recurrent division by odd or prime numbers is done. Just this strange integral:
1/2π ∫_(-π)^π▒〖e^pjω (∑_(k=2)^(k=m+1)▒〖e^(-2kjω) ((1-e^(-mkjω))/(1-e^(-kjω) )) 〗)dω 〗
Being p the number that we want to check if it is a prime or not. And being m whatever integer number higher than (p+1)/2(the lowest, the most efficient the operation). As k and ω are independent variables, the sum can be taken outside the integral (as it is in the above program).
To get to this point, we will do the following. First, we will create a domain with all the composite numbers. This is easy, as you can just multiply one by one all the integers (greater or equal than 2) in that domain. So, you will get all the composite numbers (not getting any prime) in that domain.
Then, we will use the Fourier transform to change from this original domain (called discrete time domain in this regards) to the frequency domain. There, we can “ask”, using Parseval’s theorem, if a certain number is there or not. The use of Parseval’s theorem leads to the above integral. If the number p that we want to check is not in the domain, the result of the integral is zero and the number is a prime. If instead, the result is an integer, this integer will tell us how many permutations of two divisors the number p has. And, in consequence information how many factors, the number p has.
So, for any number p lower than 2m-1, you can check if it is prime or not, just making the numerical definite integration (but even this integral, if no further developments are done, the numerical integration is inefficient computing-wise compared with brute-force checking for example). To be added, is the question regarding the level of accuracy needed (number of decimals and number of steps in the numerical integration) to have a reliable result for large numbers. This will be commented on the paper, but a separate study will be needed to have detailed conclusions.
Of course, the best would be that in the future, an analytical result (or at least an approximation) for the summation or for the integration is achieved.

**Category:** Number Theory

[937] **viXra:1812.0340 [pdf]**
*replaced on 2018-12-21 15:57:08*

**Authors:** Wu ShengPing

**Comments:** 4 Pages.

The main idea of this article is simply calculating integer
functions in module. The algebraic in the integer modules is studied in
completely new style. By a careful construction the result that
two finite numbers is with unequal logarithms in a corresponding module is proven, which result is applied to solving
a kind of high degree diophantine equation.

**Category:** Number Theory

[936] **viXra:1812.0312 [pdf]**
*replaced on 2018-12-19 06:26:14*

**Authors:** Julian Beauchamp

**Comments:** 3 Pages.

In 2002 Preda Mihailescu used the theory of cyclotomic fields and Galois modules to prove Catalan's Conjecture. In this short paper, we give a very simple proof. We first prove that no solutions exist for a^x-b^y=1 for a,b>0 and x,y>2. Then we prove that when x=2 the only solution for a is a=3 and the only solution for y is y=3.

**Category:** Number Theory

[935] **viXra:1812.0305 [pdf]**
*replaced on 2018-12-25 04:29:58*

**Authors:** Juan Moreno Borrallo

**Comments:** 24 Pages.

In this paper it is proposed and proved an exact formula for the prime-counting function, finding an expression of Legendre's formula. As corollaries, they are proved some important conjectures regarding prime numbers distribution.

**Category:** Number Theory

[934] **viXra:1812.0242 [pdf]**
*replaced on 2018-12-14 12:37:47*

**Authors:** Jean Pierre Morvan

**Comments:** 6 Pages.

L'hypothèse de RIEMANN est fausse parce que la fonction zéta n'est pas nulle avec s = 1/2 + ib

**Category:** Number Theory

[933] **viXra:1812.0235 [pdf]**
*replaced on 2018-12-16 19:13:27*

**Authors:** Toshiro Takami

**Comments:** 9 Pages.

Based on whether or not the remainder divided by 30 and 48 is a prime number, the prime number was judged.
It was difficult to judge which one was better, it seemed better to use both.
And, I was searching for a prime number determination method to be a multiple of 48, and to be able to judge with a remainder of multiples of 30, I came across a kind of pseudo prime number 203.

**Category:** Number Theory

[932] **viXra:1812.0208 [pdf]**
*replaced on 2019-01-23 08:21:08*

**Authors:** Kenneth A. Watanabe

**Comments:** 14 Pages. Acknowledgement section was added and proof of delta pi was added

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. A more general conjecture by de Polignac states that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, a function is derived that corresponds to the number of twin primes less than n for large values of n. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using this same methodology, the de Polignac Conjecture is also shown to be true.

**Category:** Number Theory

[931] **viXra:1812.0208 [pdf]**
*replaced on 2019-01-09 08:19:00*

**Authors:** Kenneth A. Watanabe

**Comments:** 13 Pages. Paper was modified to include future directions and additional references were added

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. A more general conjecture by de Polignac states that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, a function is derived that corresponds to the number of twin primes less than n for large values of n. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using this same methodology, the de Polignac Conjecture is also shown to be true.

**Category:** Number Theory

[930] **viXra:1812.0208 [pdf]**
*replaced on 2018-12-27 09:47:28*

**Authors:** Kenneth A. Watanabe

**Comments:** 12 Pages.

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. The first mention of the Twin Prime Conjecture was in 1849, when de Polignac made the more general conjecture that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, I derive a function that corresponds to the number of twin primes less than n for large values of n. This equation is identical to that used to prove the Goldbach Conjecture. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using the same methodology, de Polignac’s conjecture is also shown to be true.

**Category:** Number Theory

[929] **viXra:1812.0182 [pdf]**
*replaced on 2018-12-30 15:25:04*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, from a,b,c positive integers relatively prime with c=a+b, we consider a bounded of c depending of a,b. Then we do a choice of K(\epsilon) and finally we obtain that the ABC conjecture is true. Four numerical examples confirm our proof.

**Category:** Number Theory

[928] **viXra:1812.0182 [pdf]**
*replaced on 2018-12-16 05:58:38*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we use the Fermat's Last Theorem (FLT) to give a proof of the ABC conjecture. We suppose that FLT is false ===> we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true ===> FLT is true. But, as FLT is true, then we deduce that the ABC conjecture is true.

**Category:** Number Theory

[927] **viXra:1812.0154 [pdf]**
*replaced on 2018-12-13 06:58:30*

**Authors:** M. Sghiar

**Comments:** 7 Pages. Accepted & french version © Copyright 2018 by M. Sghiar. All rights reserved. Respond to the author by email at: msghiar21@gmail.com

I show here that if $ x \in \mathbb{N}^*$ then $1 \in \mathcal{O}_S (x)= \{ S^n(x), n \in \mathbb{N}^* \} $ where $ \mathcal{O}_S (x)$ is the orbit of the function S defined on $\mathbb{R}^+$ by $S(x)= \frac{x}{2} + (x+\frac{1}{2}) sin^2(x\frac{\pi}{2})$, and I deduce the proof of the Syracuse conjecture.

**Category:** Number Theory

[926] **viXra:1812.0150 [pdf]**
*replaced on 2019-02-18 09:09:13*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null set $\{(s,t,u)|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null set $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[925] **viXra:1812.0150 [pdf]**
*replaced on 2019-02-13 19:33:18*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null set $\{(s,t,u)|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null set $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[924] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-31 20:27:12*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null sets $\{(s,t,u)|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null sets $\{z,y,x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[923] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-29 20:38:53*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $t$; and, we \emph{denote} $2^\frac{2}{n}q$ by $u$. For any given $n>2$ : Since the term $u$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null sets $\{(s,t,u)|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null sets $\{z,y,x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. We show it is true, for $n>0$, that $\{u|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}=\{x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[922] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-22 19:55:05*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $t$; and, we \emph{denote} $2^\frac{2}{n}q$ by $u$. For any given $n>2$ : Since the term $u$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null sets $\{(s,t,u)|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null sets $\{z,y,x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. We show it is true, for $n>0$, that $\{u|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}=\{x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[921] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-19 20:14:58*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ for which $n$ is any given positive natural number, $r$ is positive real and $q$ is positive rational such that the set of triples $\{((r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},2^\frac{2}{n}q)\}$ is not empty with $(r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},(2^\frac{2}{n}q)\in\mathbb{N}$. We relate this set of triples to $\{z,y,x|z,y,x\in\mathbb{N}$ for which the transposed \emph{Fermat equation} $z^n-y^n=x^n$ holds. We demonstrate, for any given value of $n$, that $2^\frac{2}{n}q=x$. Clearly, for $n>2$, the term $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational. Consequently, for values of $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[920] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-14 21:02:12*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ for which $n$ is any given positive natural number, $r$ is positive real and $q$ is positive rational such that the set of triples $\{((r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},2^\frac{2}{n}q)\}$ is not empty with $(r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},(2^\frac{2}{n}q)\in\mathbb{N}$. We relate this set of triples to $\{z,y,x|z,y,x\in\mathbb{N}$ for which the transposed \emph{Fermat equation} $z^n-y^n=x^n$ holds. We demonstrate, for any given value of $n$, that $2^\frac{2}{n}q=x$. Clearly, for $n>2$, the term $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational. Consequently, for values of $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[919] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-09 21:44:36*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT simply for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not by way of contradiction) of FLT is based on our algebraic identity $((r^n+2q^n)^\frac{1}{n})^n-((r^n-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ such that $n$ is any given positive natural number, $r$ is unrestricted positive real and $q$ are all positive rationals, for which the set of triples $\{((r^n+2q^n)^\frac{1}{n},(r^n-2q^n)^\frac{1}{n},2^\frac{2}{n}q\}$ is not empty with $(r^n+2q^n)^\frac{1}{n},(r^n-2q^n)^\frac{1}{n},(2^\frac{2}{n}q)\in\mathbb{N}$. We relate this identity to the transposed \emph{Fermat equation} $z^n-y^n=x^n$ for which $z,y,x$ are such natural numbers. We demonstrate, for any given value of $n$, that $2^\frac{2}{n}q=x$. Clearly, for $n>2$, the term $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational. Consequently, for values of $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[918] **viXra:1812.0150 [pdf]**
*replaced on 2018-12-11 10:03:58*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT simply for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not by way of contradiction) of FLT is based on our algebraic identity, denoted, {for convenience}, as $(r)^n+(s)^n=(t)^n$ with $r,s,t>0$ as functions of variables. We infer that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n+(t)^n\}=\{(x,y,z)|x,y,z\in\mathbb{N},(x)^n+y^n=z^n\}$ for $n\in\mathbb{N}, n>2$. In addition, we show, for integral values of $n>2$, that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n=t^n\}=\varnothing$. Hence, for $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[917] **viXra:1812.0074 [pdf]**
*replaced on 2018-12-10 15:13:28*

**Authors:** Stephen Marshall

**Comments:** 8 Pages.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zero’s only at the negative even integers and complex numbers with real part 1/2
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000).
It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.
The Riemann hypothesis implies significant results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann zeta function is defined for complex s with real part greater than 1 by the absolutely convergent infinite series:
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
The Riemann hypothesis asserts that all interesting solutions of the equation:
ζ(s) = 0
lie on a certain vertical straight line.
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn = 1 + 1/2+1/3+1/4+⋯+ 1/n = ∑_(n=1)^n▒1/n
Harmonic numbers have been studied since early times and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function.
The harmonic numbers roughly approximate the natural logarithm function and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers.

**Category:** Number Theory

[916] **viXra:1812.0074 [pdf]**
*replaced on 2018-12-06 08:38:16*

**Authors:** stephen Marshall

**Comments:** 7 Pages.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1/2
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000).
It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann zeta function is defined for complex s with real part greater than 1 by the absolutely convergent infinite series:
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
The Riemann hypothesis asserts that all interesting solutions of the equation:
ζ(s) = 0
lie on a certain vertical straight line.
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn = 1 + 1/2+1/3+1/4+⋯+ 1/n = ∑_(n=1)^n▒1/n
Harmonic numbers have been studied since early times and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function.
The harmonic numbers roughly approximate the natural logarithm function and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers.

**Category:** Number Theory

[915] **viXra:1812.0040 [pdf]**
*replaced on 2018-12-20 01:47:08*

**Authors:** Toshiro Takami

**Comments:** 5 Pages.

sqrt(30a-11)=t and sqrt(16a-7)=t are of course an expression derived from sqrt(30a+1)=t and sqrt(16a+1)=t, but decided to announce sqrt(30a-11)=t
and sqrt(16a-7)=t, because it has a feeling of producing more prime than sqrt(30a+1)=t and sqrt(16a+1)=t.
These have the advantage that they do not produce numbers that end with 5 and It is difficult to produce a multiplication of prime numbers.

**Category:** Number Theory

[914] **viXra:1812.0022 [pdf]**
*replaced on 2018-12-25 01:58:45*

**Authors:** Pankaj Mani

**Comments:** 12 Pages.

Riemann Hypothesis is TRUE if we look at the Functional Equation satisfied by the Riemann Zeta function upon analytical continuation in Game Perspective way as visualized by David Hilbert. It uses technical game theoretical concepts e.g. Nash Equilibrium to assert that Riemann Hypothesis has to be True. Needs to be looked at the Foundational Principles underlying Mathematics. In other words,it’s the game of arranging Zeros in the complex plane using the functional equation.

**Category:** Number Theory

[913] **viXra:1812.0019 [pdf]**
*replaced on 2018-12-13 09:03:08*

**Authors:** Kamal Barghout

**Comments:** 16 Pages. The material in this article is copyrighted. Please obtain authorization from the author before the use of any part of the manuscript

A probabilistic proof of the Collatz conjecture is described via identifying a sequential permutation of even natural numbers by divisions by 2 that follows a recurrent pattern of the form x,1,x,1…, where x represents divisions by 2 more than once. The sequence presents a probability of 50:50 of division by 2 more than once as opposed to divisions by 2 once over the even natural numbers. The sequence also gives the same 50:50 probability of consecutive Collatz even elements when counted for division by 2 more than once as opposed to division by 2 once and a ratio of 3:1 of high counts. Considering Collatz function producing random numbers and over sufficient iterations, this probability distribution produces numbers in descending order that lead to the convergence of the Collatz function to 1, assuming the only cycle of the function is 1-4-2-1.

**Category:** Number Theory

[912] **viXra:1812.0019 [pdf]**
*replaced on 2018-12-08 04:52:18*

**Authors:** Kamal Barghout

**Comments:** 12 Pages. The material in this article is copyrighted. Please obtain authorization from the author before the use of any part of the manuscript

A probabilistic proof of the Collatz conjecture is described via identifying a sequential permutation of even natural numbers by divisions by 2 that follows a recurrent pattern of the form x,1,x,1…, where x represents divisions by 2 more than once. The sequence presents a probability of 50:50 of division by 2 more than once as opposed to divisions by 2 once over the even natural numbers. The sequence also gives the same 50:50 probability of consecutive Collatz even elements when counted for division by 2 more than once as opposed to division by 2 once and a ratio of 3:1 of high counts. Considering Collatz function producing random numbers and over sufficient iterations, this probability distribution produces numbers in descending order that lead to the convergence of the Collatz function to 1, assuming the only cycle of the function is 1-4-2-1.

**Category:** Number Theory

[911] **viXra:1812.0018 [pdf]**
*replaced on 2019-01-08 21:11:56*

**Authors:** Khalid Ibrahim

**Comments:** 96 Pages.

In this paper, not only did we disprove the Riemann Hypothesis (RH) but we also showed that zeros of the Riemann zeta function $\zeta (s)$ can be found arbitrary close to the line $\Re (s) =1$. Our method to reach this conclusion is based on analyzing the fine behavior of the partial sum of the Dirichlet series with the Mobius function $M (s) = \sum_n \mu (n) /n^s$ defined over $p_r$ rough numbers (i.e. numbers that have only prime factors greater than or equal to $p_r$). Two methods to analyze the partial sum fine behavior are presented and compared. The first one is based on establishing a connection between the Dirichlet series with the Mobius function $M (s) $ and a functional representation of the zeta function $\zeta (s)$ in terms of its partial Euler product. Complex analysis methods (specifically, Fourier and Laplace transforms) were then used to analyze the fine behavior of partial sum of the Dirichlet series. The second method to estimate the fine behavior of partial sum was based on integration methods to add the different co-prime partial sum terms with prime numbers greater than or equal to $p_r$. Comparing the results of these two methods leads to a contradiction when we assume that $\zeta (s)$ has no zeros for $\Re (s) > c$ and $c <1$.

**Category:** Number Theory

[910] **viXra:1812.0018 [pdf]**
*replaced on 2018-12-22 01:07:51*

**Authors:** Khalid Ibrahim

**Comments:** 95 Pages.

In this paper, not only did we disprove the Riemann Hypothesis (RH) but we also showed that zeros of the Riemann zeta function $\zeta (s)$ can be found arbitrary close to the line $\Re (s) =1$. Our method to reach this conclusion is based on analyzing the fine behavior of the partial sum of the Dirichlet series with the Mobius function $M (s) = \sum_n \mu (n) /n^s$ defined over $p_r$ rough numbers (i.e. numbers that have only prime factors greater than or equal to $p_r$). Two methods to analyze the partial sum fine behavior are presented and compared. The first one is based on establishing a connection between the Dirichlet series with the Mobius function $M (s) $ and a functional representation of the zeta function $\zeta (s)$ in terms of its partial Euler product. Complex analysis methods (specifically, Fourier and Laplace transforms) were then used to analyze the fine behavior of partial sum of the Dirichlet series. The second method to estimate the fine behavior of partial sum was based on integration methods to add the different co-prime partial sum terms with prime numbers greater than or equal to $p_r$. Comparing the results of these two methods leads to a contradiction when we assume that $\zeta (s)$ has no zeros for $\Re (s) > c$ and $c <1$.

**Category:** Number Theory

[909] **viXra:1811.0501 [pdf]**
*replaced on 2018-12-06 16:08:33*

**Authors:** Toshiro Takami

**Comments:** 4 Pages.

It is at a glance that pairs that can be thought of as twin prime exist at non-trivial zero points.
I considered it.

**Category:** Number Theory

[908] **viXra:1811.0338 [pdf]**
*replaced on 2018-11-23 22:07:51*

**Authors:** Marco Ripà

**Comments:** 3 Pages.

We provide a preliminary proof of Ripà’s Conjecture 3 about the convergence speed of tetration, published in October 2018, which states that, for any natural number "v", exists (at least) another natural number "a", not a multiple of 10, such that V(a)=v, where V(a) represents the convergence speed of the tetration a^^b.

**Category:** Number Theory

[907] **viXra:1811.0145 [pdf]**
*replaced on 2018-11-19 03:58:47*

**Authors:** Toshiro Takami

**Comments:** 9 Pages.

I tried variously.
(30a+bi)^2+k
(24a+bi)^2+k
(1007a+bi)^2+k
(60a+bi)^2+k etc.
(a, b and k are positive integer.)
Only the real part of the complex number was extracted.
However, in the above formula it did not work well.
and, It settled down.
(√24a+4i)^2+33
and
(√6a+4i)^2+33
I half successful.
√8, √12, √14, √18 did not succeed.
Last,
(√10a+4i)^2+35
(a are positive integer)
I half successful.
Only the real part of the complex number was extracted.
However, a relatively large number of things that are not prime numbers are still included.
The challenge to my prime production ceremony will continue.

**Category:** Number Theory

[906] **viXra:1811.0116 [pdf]**
*replaced on 2018-12-22 21:36:55*

**Authors:** Colin James III

**Comments:** 1 Page.

Properties of the zeta function of the Riemann hypothesis are not confirmed as tautologous and hence refute it.

**Category:** Number Theory

[905] **viXra:1811.0112 [pdf]**
*replaced on 2018-11-09 19:22:45*

**Authors:** Es-said En-naoui

**Comments:** 5 Pages.

The Goldbach conjecture dates back to 1742 ; we refer the reader to [1]-[2] for a history of the conjecture. Christian Goldbach stated that every odd integer greater than seven can be written as the sum of at most three prime numbers. Leonhard Euler then made a stronger conjecture that every even integer greater than four can be written as the sum of two primes. Since then, no one has been able to prove the Strong Goldbach Conjecture.\\
The only best known result so far is that of Chen [3], proving that every sufficiently large even integer N can be written as the sum of a prime number and the product of at most two prime numbers. Additionally, the conjecture has been verified to be true for all even integers up to $4.10^{18}$ in 2014 , J\"erg [4] and Tom\'as [5]. In this paper, we prove that the conjecture is true for all even integers greater than 8.

**Category:** Number Theory

[904] **viXra:1811.0080 [pdf]**
*replaced on 2019-01-14 10:19:18*

**Authors:** Ralf Wüsthofen

**Comments:** 1 Page.

This note proves the inconsistency of the Peano arithmetic (PA) by deriving both a strengthened form of the strong Goldbach conjecture and its negation.

**Category:** Number Theory

[903] **viXra:1811.0080 [pdf]**
*replaced on 2019-01-06 08:29:08*

**Authors:** Ralf Wüsthofen

**Comments:** 1 Page.

This note proves the inconsistency of the Peano arithmetic (PA) by deriving both a strengthened form of the strong Goldbach conjecture and its negation.

**Category:** Number Theory

[902] **viXra:1811.0080 [pdf]**
*replaced on 2018-12-25 10:22:01*

**Authors:** Ralf Wüsthofen

**Comments:** 1 Page.

This note proves the inconsistency of the Peano arithmetic (PA) by deriving both a strengthened form of the strong Goldbach conjecture and its negation.

**Category:** Number Theory

[901] **viXra:1811.0080 [pdf]**
*replaced on 2018-12-20 20:12:08*

**Authors:** Ralf Wüsthofen

**Comments:** 1 Page.

**Category:** Number Theory

[900] **viXra:1811.0080 [pdf]**
*replaced on 2018-12-09 18:46:57*

**Authors:** Ralf Wüsthofen

**Comments:** 1 Page.

**Category:** Number Theory

[899] **viXra:1811.0072 [pdf]**
*replaced on 2019-01-01 23:58:16*

**Authors:** Quang Nguyen Van

**Comments:** 4 Pages.

We give a expression of w^n and the possible to apply for solving Fermat's Last theorem.

**Category:** Number Theory

[898] **viXra:1811.0032 [pdf]**
*replaced on 2018-11-04 14:37:02*

**Authors:** Hassine Saidane

**Comments:** 4 Pages.

Abstract. Optimization of relevant concepts such as action or utility functions enabled the derivation of theories and laws in some scientific fields such as physics and economics. This fact suggested that the problem of the location of the Riemann Zeta Function’s (RZF) nontrivial zeros can be addressed in a mathematical programming framework. Using a constrained nonlinear optimization formulation of the problem, we prove that RZF’s nontrivial zeros are located on the critical line, thereby confirming the Riemann Hypothesis. This result is a direct implication of the Kuhn-Tucker necessary optimality conditions for the formulated nonlinear program.
Keywords: Riemann Zeta function, Riemann Hypothesis, Optimization, Kuhn-Tucker conditions.

**Category:** Number Theory