Looking for better proof to Rutgers Algebra Qual Problem On the August 2016 Rutgers Qualifying Exam appears the following problem:
Let $B$ be a symmetric nondegenerate form on a the 2 dimensional vector space over a field $\mathbb{F}_p$, where $p$ is an odd prime. Show that there is a vector $v$ such that $B(v,v)=1$.
The way I solved this (kinda brute force and did not use symmetric condition) was as follows:
Let $a=B(e_1,e_1), b=B(e_1,e_2), c=B(e_2,e_1), d=B(e_2,e_2)$ (where nondegeneracy gives $a,b=0$
Then if $v=xe_1+ye_2$, $B(v,v)=ax^2+(b+c)xy+dy^2$, so $B(v,v)=1$ can be restated as:
$(x+\frac{y(b+c)}{2a})^2 =\frac{1}{a}-\frac{d}{a}y^2$
Now set $\alpha=\frac{1}{a}, \beta=\frac{-d}{a}$ and note $\alpha,\beta\ne 0$
A solution $x,y$ exists in case $\alpha+\beta y^2$ is a quadratic residue mod $p$.
Now note that when $y$ varies from $0$ to $p-1$, $y^2$ repeats each value exactly once excepts $0$, so $\alpha+\beta y^2$ takes on $\frac{p+1}{2}$ values, but there are only $\frac{p-1}{2}$ non-quadriatic residues (mod p).
I don't like this solution because I never used the symmetry of the form, and don't think it was intended to be solved as a number theory problem.
Does anyone know of a more natural algebraic approach?
 A: More natural algebraic approach:
Suppose to the contrary that there exists no such vector $v$. For the vector space $V$ (over $F_p$) let $e_1$ and $e_2$ denote the basis vectors. Next, consider the cyclic subspaces generated by $e_1$ and $e_2$, denoted $<e_1>$ and $<e_2>$ respectively. The cardinalities of both of these subspaces is $p$ (obviously). Now, given the symmetric bilinear form $B$, suppose $B(v,v) \neq 1$ and $B(u,u) \neq 1$ for all $ v \in <e_1>$ and $ u \in <e_2>$. Since the field $F_p$ has $p$ elements and there are $p$ elements in both $<e_1>$ and $<e_2>$, there exists $ae_1, be_1 \in <e_1>$ and $ce_2, de_2 \in <e_2>$ such that $B(ae_1, ae_1)=B(be_1,be_1)$ and $B(ce_2,ce_2)=B(de_2,de_2)$ ($ a \neq b, c \neq d$) (by the pigeonhole principle.) Now, this is the case if and only if (1) $B(e_1(a-b), e_1(a-b))=0$ and  (2) $B(e_2(c-d), e_2(c-d))=0$. Given this, I now claim that $B(e_1(a-b) +e_2(c-d),x)=0$ for all $ x \in V$. Let $x \in V$. Then we have $x=\alpha e_1 + \beta e_2$, for $\alpha , \beta \in F_p$. Since $F_p$ is a field,  there exists $m_1$, $m_2$ such that $m_1 (a-b)=\alpha$ and $m_2(c-d)=\beta$. Now, multiply (1) and (2) by $m_1$ and $m_2$ respectively to get, (3) $B(e_1(a-b),\alpha e_1)$ and (4) $B(e_2(c-d), \beta e_2)$. Adding (3) and (4) we get $B(e_1(a-b) +e_2(c-d), \alpha e_1 + \beta e_2)=B(e_1(a-b) +e_2(c-d), x)=0$. This violates the non-degeneracy hypothesis.
Hope this works!
