perplexus.info :: Just Math : Positive Integer Value Poser

Positive Integer Value Poser (Posted on 2015-11-22)

Determine all possible values of a positive integer M such that the equation:
x^M + (x+2)^M + (2-x)^M = 0 has at least one integer solution.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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solution

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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

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