On a Few Diophantine Equations Related to Fermat’s Last Theorem

2002 ◽  
Vol 45 (2) ◽  
pp. 247-256 ◽  
Author(s):  
O. Kihel ◽  
C. Levesque

AbstractWe combine the deep methods of Frey, Ribet, Serre and Wiles with some results of Darmon, Merel and Poonen to solve certain explicit diophantine equations. In particular, we prove that the area of a primitive Pythagorean triangle is never a perfect power, and that each of the equations X4−4Y4 = Zp, X4 + 4Yp = Z2 has no non-trivial solution. Proofs are short and rest heavily on results whose proofs required Wiles’ deep machinery.

2003 ◽  
Vol 2003 (71) ◽  
pp. 4473-4500
Author(s):  
C. Levesque

This is a survey on Diophantine equations, with the purpose being to give the flavour of some known results on the subject and to describe a few open problems. We will come across Fermat's last theorem and its proof by Andrew Wiles using the modularity of elliptic curves, and we will exhibit other Diophantine equations which were solvedà laWiles. We will exhibit many families of Thue equations, for which Baker's linear forms in logarithms and the knowledge of the unit groups of certain families of number fields prove useful for finding all the integral solutions. One of the most difficult conjecture in number theory, namely, theABC conjecture, will also be described. We will conclude by explaining in elementary terms the notion of modularity of an elliptic curve.


Author(s):  
Benson Schaeffer

In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem. AMS 2020 subject classification: 11A99, 11D41 Diophantine equations, Fermat’s equation ∗The corresponding author. E-mail: [email protected] 1 1 Introduction To prove Fermat’s Last Theorem, it suffices to show that the equation A p + B p = C p (1In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem.


Author(s):  
Robin Wilson

‘More triangles and squares’ explores Diophantine equations, named after the mathematician Diophantus of Alexandria. These are equations requiring whole number solutions. Which numbers can be written as the sum of two perfect squares? Joseph-Louis Lagrange’s theorem guarantees that every number can be written as the sum of four squares, and Edward Waring correctly suggested that there are similar results for higher powers. In 1637, Fermat conjectured that no three positive integers, a, b, and c, can satisfy the equation an+bn=cn, if n is greater than 2. Known as ‘Fermat’s last theorem’, this conjecture was eventually proved by Andrew Wiles in 1995.


2019 ◽  
Author(s):  
OLIVIER IDRISS 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} &amp; : \mathbb{N}^{3}&amp; &amp;\to &amp; \mathbb{Z}\\&amp; &amp; (X,Y,Z) &amp; \mapsto &amp; 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


1986 ◽  
Vol 59 (2) ◽  
pp. 76 ◽  
Author(s):  
Jonathan P. Dowling

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