How many solutions which integers such that x^2 + y^2 + z^2 = 2016

Question

ganeshie8 · Answer

I think it is sufficient to consider solutions in positive integers first, then we can build the complete solutions by permutations.

ganeshie8 · Answer

Working mod 4 it is easy to see that each of x, y, z must be of form $4k$. Otherwise left hand side is not divisible by 16. So the equation reduces to 
$$a^2+b^2+c^2 = 126$$
where $x = 4a$, $y=4b$, $z=4c$

anonymous · Answer

apparently gauss did it over 200 years ago but hmm

ganeshie8 · Answer

related to quadratic reciprocity law in any ways ?
maybe taking mod 7 or 3 might narrow it further idk... my attempt above looks very hacky..

anonymous · Answer

actually, he did it for square free numbers, so I guess we're screwed

ganeshie8 · Answer

looks interesting http://gyazo.com/301173b00539b4577e227beafd0b4281

dan815 · Answer

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