A Different Look at the Primitive Integral Triads of z n = y n + xn (n = 2) and a Conjecture on z n - xn for any n(¹ 2)

Abstract

The primitive Pythagorean triples (x, y, z) are now well understood [1]. However, we believe that a closer look at the solution is needed along new directions to understand the terrible difficulty in giving a simple proof for the Fermat’s last theorem. Keeping this fact in mind we look at the solutions of z 2 = y 2 + x2 , (x, y) = 1 in the following manner. (x, y, z) is a primitive Pythagorean triple if and only if x2 + y 2 = z 2 , (x, y) = 1 (1)It is obvious that one of (x, y, z) is even and it can be shown that z is never even by using (1) and substituting z = y + p, p ³ 1, in it. Now either x or y is even. If we suppose that y is even, z 2 - x2 = y 2 and then it follows that z - x = 22b -1 or z - x = 22b -1a 2 where a ,b ³ 1 and are integers. The following are examples for the justification of our point. 2 2 2 3 2 2 2 13 12 5 , 2 , 2 17 15 8 , 1 , 2 = + = - = = + = - = z x z x b b 1132 = 1122 +152 ,b = 1,a = 7 z - x = 2´ 72 Now we apply the mean value theorem of the form a2 - b2 = 2(a - b)x where a <x < b , to the expression z 2 - x2 , to obtain z 2 - x2 = 2.22b -1 a 2x since z 2 - x2 = (z - x)(z + x) It follows that ( )( ) 2 2 2 2. z x z x z x - = - + It is clear that 2.22b -1a 2 or 2(z - x) is a perfect square and since ( )( ) 2 2 2. z x y z x = - + it follows that =x + 2 z x is a perfect square. Therefore, in case of any primitive triple (x, y, z) of z 2 = y 2 + x2 , the mean value theorem is manifested in the form a2 - b2 = 2(a - b)x where x is a perfect square b <x < a . Now we point out the following conjecture. Suppose that z, x > n for any prime n ³ 3 . Then, z n - xn = n (z - x)x n-1 by the mean value theorem and we conjecture that x is irrational when z - x =a nnbn-1 .