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): Tricks or Treats? by ed bottemiller)

RE comment about "Hop, Skip and Jump",

"By the way, I enjoyed your jumping game; the best I found was the 321321321 which I encoded in my comment -- for which my apologies if I gave offense.  ADJNLHCMNR somehow lacked in euphony."

I would hardly think there would be any cause for Charlie to take offence, very frequently members in the past have voiced solutions
which in reality have been spoilers but they have been disguised, sometimes more subtlely than was yours.  Such an answer does not inhibit someone who, feeling a little reticent, may feel they have a solution worthy of offer.

Posted by brianjn on 2010-11-05 01:41:45