Given an equation x^2+y^2+z^2=2010.
1. Prove that in every triplet of integers satisfying the above equation one number has to be even and two others odd.
2. How many integer solutions are there ?


Warning: 1 is very easy, 2 is quite tricky.

(In reply to re(2): solutions by Charlie)

Yes, I now see that, thanks!

Posted by broll on 2010-11-05 11:40:46