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Show that in any finite field, there exists a non-trivial solution for $x^2 + y^2 + z^2 + w^2 = 0$.

I have shown it for finite fields of cardinality $q$, when $4$ divides $q-1$ and when $q-1$ is odd. In such finite fields, $-1$ has a square root. Hence, $x^2+y^2 = 0$ has a non-trivial solution.

I want a non-trivial solution for the above equation when $q-1 \equiv 2 (\mod 4)$? Here, we already know that there is no non-trivial solution for $x^2+y^2 = 0$.

Is there a more general approach that does not care about the cardinality of the field?

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8  
Apply Lagrange's four-square theorem to the characteristic of the field? –  achille hui Sep 19 '13 at 11:38
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Actually a simple counting argument shows that there are always solutions with $z=1$ and $w=0$. It's been done in this site many times. Search for it. –  Jyrki Lahtonen Sep 19 '13 at 11:41
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Hmm. Couldn't find it. Basically the set of elements of the form $1+y^2$ has $(q+1)/2$ elements. The same holds for the set of elements of the form $-x^2$. There those two sets must intersect non-trivially. –  Jyrki Lahtonen Sep 19 '13 at 11:49
    
Thank you @JyrkiLahtonen. –  Arpita Korwar Sep 19 '13 at 12:31

1 Answer 1

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As Jyrki points out, the set of elements of the form $1+y^2$ has $(q+1)/2$ elements where $q$ is the cardinality of the field. Similarly for set of elements of the form $-x^2$. See my answer for proofs of these facts.

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