On The Exceptional solutions of Jeśmanowicz' conjecture

Abstract

Let (a, b, c) be a primitive Pythagorean triple. Set a = m 2 −n 2 , b = 2mn , and c = m 2 + n 2 with m and n positive coprime integers, m > n and m ≡ n (mod 2). A famous conjecture of Jeśmanowicz asserts that the only positive integer solution to the Diophantine equation a x + b y = c z is (x, y, z) = (2, 2, 2). In this note, we will prove that for any n > 0 there exists an explicit constant c(n) such that if m > c(n), the above equation has no exceptional solution when all x,y and z are even. Our result improves that of Fu and Yang [11]. As an application, we will show that if 4 || m and m > C(n), then Jeśmanowicz' conjecture holds.c r with some conditions on p,q, and r, then disapprove the possibility of the existence of any other solution.There are much work, results, and conjectures by several authors in this direction (e.g., [34], [5] , [6]...), which lead to the following conjecture [21]:Conjecture 1.1. For given coprime integers a, b, c > 1, the Diophantine equation (1.1) has at most one solution in integers x, y, z > 1.The most famous, old example is when a, b, and c are primitive Pythagorean triple, and hence p = q = r = 2 is a solution. Sierpiński [33] showed that the equation 3 x + 4 y = 5 z has a unique solution (2, 2, 2) in positive integers. In the same year, Jeśmanowicz [18] proved the same result for the Pythagorean triple (a, b, c) = (5, 12, 13), (7, 24, 25), (9, 40, 41) and (11; 60; 61). He further made the following conjecture: Conjecture 1.2. Let (a, b.c) be positive integers where a 2 +b 2 = c 2 and GCD(a, b) = 1. Then the Diophantine equation a x + b y = c z has only the positive integer solution (x, y, z) = (2, 2, 2).