## Number Theory   ## Discovery on Beal Conjecture

Authors: Idriss Olivier 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

Comments: 7 Pages.

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### Submission history

[v1] 2019-04-06 00:57:16

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