M can't be even because then we have the sum of three squares = 0 and only one of the squares can equal zero.
If M=1, x + (x+2) + (2-x) = 0 and x = -4 is a solution.
For other odd M and -2<=x<=2 there are no solutions.
For odd M with x outside that range we can rearrange the equation to read A^M + B^M = C^M with each of (A,B,C) positive, which we know thanks to Andrew Wiles is an impossibility.
Thus M=1 is the only solution.
Posted by xdog
on 2015-11-23 15:35:46