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Let $V=V_{n}(q)$ be a $n$ dimensional vector space over the finite field $\mathbb{F}_{q}$ and let $(,)$ be a symmetric bilinear form on $V$. Fix $v\in V$. I would like to show that there exists a vector $w\in \langle v\rangle^{\bot}$ such that $(v,v).(w,w)$ is a square in $\mathbb{F}_{q}$.

I'm not sure why such a $w$ must exist?

I would really appreciate any help. Thanks!

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Presumably zero is not a square for you for otherwise $w=0$ works :-)

In general the claim is false. Consider for example the case $q=3$, $n=2$ and the symmetric bilinear form $$ ((x_1,x_2),(y_1,y_2))=2x_1y_1+x_2y_2. $$ If $v=(1,0)$, then $(v,v)=2$ is a non-square. The orthogonal complement $\langle v\rangle^\perp$ is spanned by $(0,1)$, so we must use $w=(0,a)$ for some $a\neq0$. But $$(w,w)=a^2,$$ so $$(v,v)(w,w)=2a^2,$$ which is a non-square for all choices of $a$.


The claim does hold if we make a few additional assumptions. I am not sure about the most general case, but I can prove it, if we assume that $(v,v)\neq0$, $n\ge3$ and that $(\ ,\ )$ is non-degenerate. Assuming these (hoping that these condition are met in the case where you want to apply this).

If $q$ is even, then all the elements of $\Bbb{F}_q$ are squares and there is nothing to prove.

Assume $q$ is odd. Everything that follows revolves about the fact that in $\Bbb{F}_q^*$ the product of two non-squares is a square, as is the product of two squares but the product of a square and a non-square is a non-square. If you are not familiar with this, you should first refresh those bits.

The non-degeneracy of $(\ ,\ )$ means that $W=\langle v\rangle^\perp$ is at least 2-dimensional. The restriction of $(\ ,\ )$ to $W$ is still non-degenerate and, because $q$ is odd, it is diagonalizable. So we can choose two linearly independent vectors $\{w_1,w_2\}$ in $W$ such that we have $$ f(x,y):=(xw_1+yw_2,xw_1+yw_2)=d_1x^2+d_2y^2 $$ for some non-zero constants $d_1,d_2\in\Bbb{F}_q$. It suffices to show that there exist points $(x,y)$ such that $f(x,y)$ is a square and also points $(x,y)$ such that $f(x,y)$ is a non-square.

  • If one of $d_1,d_2$ is a square and the other a non-square, then the same holds for $f(1,0)$ and $f(0,1)$.
  • If both of $d_1,d_2$ are squares, then after rescaling the basis vector we can assume that $d_1=d_2=1$. There are $(q-1)/2$ non-zero squares, forming a subset $Q\in\Bbb{F}_q$. So if $a\in\Bbb{F}_q$ is a non-square, then the set $a-Q=\{a-x\mid x\in Q\}$ has $(q-1)/2$ non-zero elements distinct from $a$. Therefore there must be a square among them, i.e. $(a-Q)\cap Q\neq\emptyset$. So there exists squares $r=x^2$ and $s=y^2$ such that $r+s=a$. But we just showed that $f(x,y)=r+s=a$ is a non-square. Of course, $f(1,0)=d_1$ is still a square.
  • If both $d_1, d_2$ are non-squares, a similar counting argument shows that a square (actually any square) in $\Bbb{F}_q$ can be written as a sum of two non-squares. For example $1=r+s$ with some non-squares $r,s$. Then $r/d_1$ and $s/d_2$ are both squares, say $x^2$ and $y^2$ respectively. It follows that $f(x,y)=1$, so $f$ also takes squares as values.
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  • $\begingroup$ Thanks for such a detailed response! However, I don't understand why $W$ is 2-dimensional; I thought that since $<v>$ is $1$-dimensional, the perpendicular space $W$ should be $n-1$-dimensional? Am I misunderstanding something here? $\endgroup$ – Ishika Feb 14 '16 at 21:29
  • $\begingroup$ No, you understood ok. That was mistake from me. I meant to say that $W$ is at least 2-dimensional. The argument still works when we diagonalize, and pick a suitable 2-d subspace. The point was to make sure that $n-1\ge2$. $\endgroup$ – Jyrki Lahtonen Feb 14 '16 at 21:55
  • $\begingroup$ Fixed now, I think. $\endgroup$ – Jyrki Lahtonen Feb 14 '16 at 21:57
  • $\begingroup$ Why do we need the restriction of $(,)$ to be non-degenerate? A symmetric bilinear form is diagonalizable anyway, isn't it? $\endgroup$ – Ishika Feb 24 '16 at 16:57
  • $\begingroup$ Yes, it is diagonalizable. But then it may be that the restriction of $(\ ,\ )$ to $W$ vanishes identically. Then you would be forced to the situation $(w,w)=0$. I wasn't sure, whether you allow that :-) $\endgroup$ – Jyrki Lahtonen Feb 24 '16 at 19:31

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