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 solutions by Charlie)

I agree with your first part and also that the equation  x^2+y^2+z^2=2010 has 384 integer solutions. However, ignoring sign changes, there seem to be 7 values of x for which there are 4 solutions {5,7,19,25,32,35,40} (i.e. 2 different y,z values plus their reversals) and 10 values of x for which there are only 2 solutions {1,4,11,16,17,23,28,31,37,44} (i.e. one solution, plus the same with y,z, reversed) giving 8*(4*7) = 224, plus 8*(10*2) = 160, producing 384 solutions in all.

Posted by broll on 2010-11-05 02:53:01