Each of x, y and z is a positive integer satisfying:
x²+y²+z² = 2014

How many solutions are there? For which triplets is x+y+z a perfect square? How many distinct values can x+y+z have?

*** Disregard permutations. For example, (p,q,r) and (p,r,q) should be treated as the same solution.

Well, on my TI-84 anyway
3,18,41
3,22,39 sum = 64=8^2
5,15,42
5,30,33
9,13,42 sum = 64=8^2
13,18,39
14,27,33
18,27,31
21,22,33

For 9 distinct solutions
The 6 distinct sums are 62, 64, 68, 70, 74, 76

Since 2014 is even but not divisible by 4, every triplet consists of two odds and one even.  This is why x+y+z is always even.

Posted by Jer on 2014-01-16 15:38:29