Proving a Diophantine equation has no solutions

Question

I'm trying to show that $7u^2=x^2+y^2+z^2$ has no solutions in $\mathbb{Z}$ when $u$ is odd. If $u$ is even, then it's simple to show that no solutions exists by looking modulo $4$. The odd case looks harder.

Answer 1

For the odd case, let $u=1+2k$; now $u^2=1+4k+4k^2 = 1+4k(k+1)$. As $k$ or $k+1$ is even, we have $u^2\equiv1\mod8$ whenever $u$ is odd, hence $7u^2\equiv7\mod8$. For the righthand side, we have a sum of three squares mod 8, that is, three numbers taken from the set $\{0,1,4\}$ and summed $\mod8$, which cannot equal 7.