Let $\mathbb{F}_q$ denote the finite field with $q$ elements and Ch$\mathbb{F}_q\neq 2$. What is the number of solutions of the quadratic equation $X_1^2+\cdots + X_r^2=0$ in $\mathbb{F}_q^m$ for $1\leq r\leq m$.
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2$\begingroup$ If $r$ is odd, at least $3$, then the number of solutions equals $q^{m-1}$. If $r$ is even, say $2s+2$ for $s$ at least $1$ (so $r\geq 4$), then the number of solutions equals $q^{m-1} + q^{m-s-2}(q-1)$. When $r$ equals $1$, the number equals $q^{m-1}$. When $r$ equals $2$, the answer depends on $q$, e.g., if $q=p$ then the answer depends on whether $p$ is congruent to $+1$ or $-1$ modulo $4$. $\endgroup$– Jason StarrCommented Feb 14, 2018 at 16:40
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1$\begingroup$ What is the role of $r$ vs. $m$? I assumed that each $X_i$ came from $\mathbb F_q$, but then we seem to be working in $\mathbb F_q^r$, not $\mathbb F_q^m$ (or else just meaninglessly multiplying our answer by $q^{m - r}$). $\endgroup$– LSpiceCommented Feb 14, 2018 at 16:43
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3$\begingroup$ @LSpice I had the same question. I decided that we are just meaninglessly multiplying the number of solutions by $q^{m-r}$. $\endgroup$– Jason StarrCommented Feb 14, 2018 at 16:49
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$\begingroup$ Thanks, @Jason Starr. Could you please give me a reference where I can get the complete proof? $\endgroup$– SinghCommented Feb 15, 2018 at 10:40
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2$\begingroup$ I'm fairly sure this is covered in Ireland & Rosen. See also Math.SE. $\endgroup$– Jyrki LahtonenCommented Feb 16, 2018 at 22:26
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1 Answer
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The answer (already given in comments, with a small misprint/mistake) is:
(a) for even $r=2k+2$, it is $q^{m-1}+(q-1)q^{m-k-2}\eta((-1)^{k+1})$, where $\eta$ is the quadratic character of $\mathbb{F}_q$, $\eta(x)=1$ if $x$ is a square, and $\eta(x)=-1$ if $x$ is not a square.
(b) for odd $r$, it is always $q^{m-1}$.
For the specific reference, see, respectively, Theorems 6.26, 6.27 in Finite Fields by Lidl and Niederreiter. The results of those theorems should be applied in the case of dimension $r$, and then multiplied by $q^{m-r}$ (choices for the last $m-r$ coordinates).
answered Feb 27, 2018 at 15:40
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$\begingroup$ Can you also include the actual answer from the comments in your answer? Comments are not meant to be permanent. $\endgroup$ Commented Feb 27, 2018 at 17:03
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$\begingroup$ @EmilJeřábek : done (in fact, there was a little misprint/mistake in the comments in the case of even $r$). $\endgroup$ Commented Feb 27, 2018 at 17:56