Prove that there does not exist a power of 2 that is expressible as the sum of three positive perfect squares.

Any perfect square is congruent to 0 or 1 mod 4.  So if the sum of three perfect squares is to be a power of 2 it must equal 1 or 2 or be a multiple of 4.  The cases 1 and 2 are ruled out since we have three *positive* squares.  So the sum must be congruent to 0 mod 4 and each square must be even, and thus the square of an even number.  Let these even numbers be 2a, 2b and 2c.  Then we have:

(2a)² + (2b)² + (2c)² = 2^k, for some integer k >= 2.  Hence:

a² + b² + c² = 2^(k-2).

Clearly we can continue in this way until the power of 2 is either 2^0 or 2^1.  But neither of these is a solution, so by contradiction no solution is possible.

This is an example of a proof by infinite descent, pioneered by Fermat.

Posted by Nick on 2010-02-28 16:22:36