Hypothesis

[Joshua Zucker]

On 9/28/07, Jonathan Post <jvospost3 at gmail.com> wrote:
> Some cleaning up to do.  Am I missing something?

You're missing something much simpler: after 3, all primes are not
divisible by 3, hence their squares (3n +/- 1)^2 = 9n^2 +/- 6n + 1 all
leave remainder 1, so the sum of three of those squares will be a
multiple of three.

The same argument works for the fourth powers as well, and indeed any
even power.

--Joshua Zucker