I have been working through an exam question about which primes are represented by quadratic forms of discriminant $-35$.
So far I have been able to show that the only two reduced forms with discriminant $-35$ are $x^2 + xy + 9y^2$ and $3x^2 + xy + 3y^2$.
Next I showed that only primes $p$ with $\left(\frac{p}{5}\right)\left(\frac{p}{7}\right) = 1$ are represented by a form with discriminant $-35$.
However now the question asks me to show that if $\left(\frac{p}{5}\right) = \left(\frac{p}{7}\right) = 1$, $p$ is represented by $x^2 + xy + 9y^2$ and if $\left(\frac{p}{5}\right) = \left(\frac{p}{7}\right) = -1$, $p$ is represented by $3x^2 + xy + 3y^2$ and I can't see how to do it.
Please can someone show me how to do this?
ADDED: it is worth your time to calculate exactly how the possible steps in Gauss reduction of a positive binary form match perfectly with matrix updates, given $Z$ take an "elementary" matrix $E$ and calculate $E^T ZE$
In the two by two case, we usually demand positive determinant, so the legal elementary matrices are $$ E = \left( \begin{array}{cc} 1 & s \\ 0 & 1 \end{array} \right) $$
$$ E = \left( \begin{array}{cc} 1 & 0 \\ s & 1 \end{array} \right) $$
$$ E = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) $$
the way you wrote it, you need quadratic reciprocity. We begin with a (positive) prime $p \neq 2,5,7$ such that $$ (-35|p) = 1.$$ That means we can find some $$ w^2 \equiv -35 \pmod p, $$ $$ w^2 + 35 \equiv 0 \pmod p . $$ Now if $w$ is even, we replace it by $p - w \equiv -w \pmod p.$ Now $w^2 + 35 \equiv 0 \pmod 4,$ together $$ w^2 + 35 \equiv 0 \pmod {4p}. $$ This means there is an integer $t$ with $$ w^2 + 35 = 4pt. $$ Finally, $$ w^2 - 4pt = -35. $$ We have constructed the binary form with discriminant $-35$ andcoefficients $$ \langle p , \; w , \; t \rangle \; . $$ Let $H$ be the two by two matrix $$ H = \left( \begin{array}{cc} 2p & w \\ w & 2t \end{array} \right) $$ Reduce this, meaning construct the matrix $P$ with integer elements and determinant $1,$ such that $G =P^T HP$ is reduced; either $$ G = \left( \begin{array}{cc} 2 & 1 \\ 1 & 18 \end{array} \right) $$ OR $$ G = \left( \begin{array}{cc} 6 & 1 \\ 1 & 6 \end{array} \right) $$ Whichever one happens, take $$ Q = P^{-1}, $$ and $Q^T G Q = H$ shows how to represent $p$ with integers.