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) + (2x) = 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 20151123 15:35:46 