IDRISS O L I V I E R BADO

In this paper we give a proof for  Beal's conjecture . Since the discovery of the proof of Fermat's last theorem  by  Andre Wiles, several questions arise on the correctness  of Beal's conjecture. By using a very rigorous method we come to the proof. Let $ \mathbb{G}=\{(x,y,z)\in \mathbb{N}^{3}: \min(x,y,z)\geq 3\}$
$\Omega_{n}=\{ p\in \mathbb{P}: p\mid n , p \nmid z^{y}-y^{z}\}$ ,
$$\mathbb{T}=\{(x,y,z)\in \mathbb{N}^{3}: x\geq 3,y\geq 3,z\geq 3\} $$
$\forall(x,y,z) \in \mathbb{T}$ consider the function  $f_{x,y,z}$ be  the  function  defined as :
$$\begin{array}{ccccc}
f_{x,y,z} & : \mathbb{N}^{3}&  &\to & \mathbb{Z}\\
& & (X,Y,Z) & \mapsto & X^{x}+Y^{y}-Z^{z}\\
\end{array}$$
Denote by $$\mathbb{E}^{x,y,z}=\{(X,Y,Z)\in  \mathbb{N}^{3}: f_{x,y,z}(X,Y,Z)=0\}$$
and $\mathbb{U}=\{(X,Y,Z)\in \mathbb{N}^{3}: \gcd(X,Y)\geq2,\gcd(X,Z)\geq2,\gcd(Y,Z)\geq2\}$
Let  $ x=\min(x,y,z)$  . The obtained result show that :if $ A^{x}+B^{y}=C^{z}$ has a solution and  $ \Omega_{A}\not=\emptyset$, $\forall p \in \Omega_{A}$ ,

$$ Q(B,C)=\sum_{j=1}^{x-1}[\binom{y}{j}B^{j}-\binom{z}{j}C^{j}]$$ has no solution  in  $(\frac{\mathbb{Z}}{p^{x}\mathbb{Z}})^{2}\setminus\{(\overline{0},\overline{0})\} $  Using this result we show that Beal's conjecture is true since $$ \bigcup_{(x,y,z)\in\mathbb{T}}\mathbb{E}^{x,y,z}\cap \mathbb{U}\not=\emptyset$$ Then $\exists (\alpha,\beta,\gamma)\in \mathbb{N}^{3}$ such that $\min(\alpha,\beta,\gamma)\leq 2$ and    $\mathbb{E}^{\alpha,\beta,\gamma}\cap \mathbb{U}=\emptyset$
The novel techniques  use for the proof can be use to solve the variety of Diophantine equations .  We provide also the solution to Beal's equation . Our proof can provide an algorithm to generate solution to Beal's equation

Goldbach's famous conjecture has always fascinated eminent mathematicians. In this paper we give a rigorous proof based on a new formulation, namely, that every even integer has a primo-raduis. Our proof is mainly based on the application of Chebotarev-Artin's theorem, Mertens' formula and the Principle exclusion-inclusion of Moivre.

In this paper we give a proof of Beal's conjecture . Since the discovery of the proof of the last theorem of fermat by Andre Wiles, several questions arise on the conjecture of Beal. By using a very rigorous method we come to the proof. Let $ \mathbb{G}=\{(x,y,z)\in \mathbb{N}^{3}: \min(x,y,z)\geq 3\}$
$\Omega_{n}=\{ p\in \mathbb{P}: p\mid n , p \nmid z^{y}-y^{z}\}$ ,
$$\mathbb{T}=\{(x,y,z)\in \mathbb{N}^{3}: x\geq 3,y\geq 3,z\geq 3\} $$
$\forall(x,y,z) \in \mathbb{T}$ consider the function  $f_{x,y,z}$ be  the  function  defined as :
$$\begin{array}{ccccc}
f_{x,y,z} & : \mathbb{N}^{3}&  &\to & \mathbb{Z}\\
& & (X,Y,Z) & \mapsto & X^{x}+Y^{y}-Z^{z}\\
\end{array}$$
Denote by $$\mathbb{E}^{x,y,z}=\{(X,Y,Z)\in  \mathbb{N}^{3}: f_{x,y,z}(X,Y,Z)=0\}$$
and $\mathbb{U}=\{(X,Y,Z)\in \mathbb{N}^{3}: \gcd(X,Y)\geq2,\gcd(X,Z)\geq2,\gcd(Y,Z)\geq2\}$
Let  $ x=\min(x,y,z)$  . The obtained result show that :if $ A^{x}+B^{y}=C^{z}$ has a solution and  $ \Omega_{A}\not=\emptyset$, $\forall p \in \Omega_{A}$ ,

$$ Q(B,C)=\sum_{j=1}^{x-1}[\binom{y}{j}B^{j}-\binom{z}{j}C^{j}]$$ has no solution  in  $(\frac{\mathbb{Z}}{p^{x}\mathbb{Z}})^{2}\setminus\{(\overline{0},\overline{0})\} $  Using this result we show that Beal's conjecture is true since $$ \bigcup_{(x,y,z)\in\mathbb{T}}\mathbb{E}^{x,y,z}\cap \mathbb{U}\not=\emptyset$$ Then $\exists (\alpha,\beta,\gamma)\in \mathbb{N}^{3}$ such that $\min(\alpha,\beta,\gamma)\leq 2$ and    $\mathbb{E}^{\alpha,\beta,\gamma}\cap \mathbb{U}=\emptyset$
The novel techniques  use for the proof can be use to solve the variety of Diophantine equations .  We provide also the solution to Beal's equation . Our proof can provide an algorithm to generate solution to Beal's equation

In this paper, we will rst establish that there are many primes p such that p+n is prime for an even integer n , by using the Chébotarev-Artin's theorem, the inclusion-exclusion principle of Moivre, Mertens formula. With these tools we get a function whose counts the number of primes p such that p + n is prime between X + n and √ X + n where X is a real number. For m = inf{n ∈ 2N * : p + n ∈ P} we deduce Polignac's conjecture. 1 Introduction In number theory, Polignac's Conjecture was introduced by Alphonse de Polignac in 1849 and states : Given even integer n, it exists an innity of consecutive prime integers with dierence n. In other words, given even integer n , it exists an innity of prime numbers p such that p + n are simultaneously consecutive prime. Our objective is to prove in this present paper this old conjecture. We propose here a proof of this conjecture by proving even deviation prime number conjecture ,which asserts the existence of innity prime integers p such that p + n is prime for any given even integer n 1.1 Principle of the Demonstration Let X ∈ R and n an even integer ,to prove even deviation prime numbers Conjecture , we will show that the interval [

In this paper, we are going to give the proof of the Goldbach conjecture by introducing the lemma which implies Goldbach conjecture. rst of all we are going to prove that the lemma implies Goldbach conjecture and in the following we are going to prove the validity of the lemma by using Chébotarev-Artin theorem's, Mertens formula and the Principle of inclusion-exclusion of Moivre.

In this paper we are going to give the proof of Goldbach conjecture by introducing the lemma which imply Goldbach conjecture. rst of all we are going to prove that the lemma imply Goldbach conjecture and in the following we are going to prove the validity of the lemma by using Chébotarev-Artin theorem's , Mertens formula and Principle of inclusion-exclusion of Moivre .

In this article we give a proof of the prime constellations conjecture . Our proof is so rigorous  in the sense that it is based on the correctness of the Chebotarev- Artin theorem. We have all along our approach introduced a class of very interesting prime which said to be $k$ - admissible thot prime . This class is interesting because it serves as a brick to generate  prime constellations of length $k $ . Our results predict that the distribution of its prime numbers is worth :$$ (1-Y_{k,\infty})\frac{n}{\ln n}$$
where $$ Y_{k,\infty}=k-\sum_{t=1}^{k}\prod_{p\in \mathbb{P},p, \nmid a_{t}}(1-\frac{1}{p^{\beta_{\theta(p)}-1}(p-1)})$$

The purpose of this article is to introduce the theory of totherian analysis in order to provide proof of the Riemann hypothesis: the concepts introduced have been so effective and we can use it to build a coherent and tangible analysis. Totherian analysis can be considered as an effective remedy in solving a lot of problems in mathematics

In this paper, we will rst establish that there are many primes p such that p+n is prime for an even integer n , by using the Chébotarev-Artin's theorem, the inclusion-exclusion principle of Moivre, Mertens formula. With these tools we get a function whose counts the number of primes p such that p + n is prime between X + n and √ X + n where X is a real number. For m = inf{n ∈ 2N * : p + n ∈ P} we deduce Polignac's conjecture. 1 Introduction In number theory, Polignac's Conjecture was introduced by Alphonse de Polignac in 1849 and states : Given even integer n, it exists an innity of consecutive prime integers with dierence n. In other words, given even integer n , it exists an innity of prime numbers p such that p + n are simultaneously consecutive prime. Our objective is to prove in this present paper this old conjecture. We propose here a proof of this conjecture by proving even deviation prime number conjecture ,which asserts the existence of innity prime integers p such that p + n is prime for any given even integer n 1.1 Principle of the Demonstration Let X ∈ R and n an even integer ,to prove even deviation prime numbers Conjecture , we will show that the interval [