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.

  • $\begingroup$ I figured it out. For the odd case look modulo $8$. $\endgroup$
    – dst
    Nov 9, 2011 at 19:51
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    $\begingroup$ You mean non-trivial solutions. And although the even case is not hard, it requires the odd case. $\endgroup$ Nov 9, 2011 at 20:25
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    $\begingroup$ @dst Please post your approach as an answer so that other users can check and upvote it. You can even accept your answer. That way, this question will appear as answered in the future. $\endgroup$
    – Srivatsan
    Nov 9, 2011 at 21:47

1 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.


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