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: Tricks or Treats? by Charlie)

I must admit that I was careless with the first part (which you have well illustrated).  For the second part, I guess we both assumed that the only tricks Ady had in mind would be if we ignored the permutations or the added factor of the signs on x, y,z.  Those strike me not so much as tricks, but as unstated specs regarding the scope of the integers and their sequencing  -- but perhaps Ady has something else in mind. The other that came to mind would be the possible issue of operator precedence (exponentiations before additions, lacking parentheses to override).

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.

 

Posted by ed bottemiller on 2010-11-04 19:49:42