a simple proof is provided for n = 1, n = 2, n = 3, n = 4 and then for n = m. For n = 1, 2 solution exists whereas for n > 2, there is no solution. In equation (1), we can write y and z as (x + a) and (x + b) respectively for some positive integers a and b such that b > a. This is without any loss of generality. Thus, equation (1) can be written as below.
xn + (x + a)n = (x + b)n ………………………………..…………………… (2)
Where x, a, b, n, are positive integers. Also for n = 1, x = (b – a) and for n 2, x > (b – a). The proof consists of two parts: existence of solution for n = 1, 2 and non-existence of solution for n > 2. It is proved in the paper that there exist positive integer x for some positive integers a, b and a < b such that equation (2) is true when n = 1, 2.
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