# A question for students you dislike

At the recent CNTA conference, Professor Joe Silverman gave an explicit homomorphism of $\text{GL}_3(\mathbb{R})$ into $\text{GL}_6(\mathbb{R})$ which he jokingly called a great question to ask undergraduates to work out explicitly… if you don’t like them very much. In a similar vein, here is a question that one might ask undergraduates they don’t particularly like:

Let $\mathbf{a} = (\alpha, \beta, \gamma)$ be a triple of co-prime integers which are not all zero and such that $\alpha \gamma > 1$. Prove that the ternary quadratic form given by:

$K_{\mathbf{a}} (F) = \dfrac{1}{8\alpha^3} \bigg( 72 \beta^2 \gamma A^2 + 9 \alpha(\beta^2 + 4 \alpha \gamma) B^2 + 8 \alpha^3 C^2 - 18\beta (\beta^2 + 4 \alpha \gamma)AB$
$+ 12 \alpha (3 \beta^2 - 4 \alpha \gamma)AC - 24 \alpha^2 \beta BC \bigg)$

takes on integer values whenever the triplet $(A,B,C)$ lies in the lattice defined by the congruence conditions

$4 \beta \gamma x - (2 \beta^2 + \alpha \gamma) y + 2 \alpha \beta z \equiv 0 \pmod{\alpha^2}$

and

$\gamma(2 \beta^2 + \alpha \gamma)x - \beta(\beta^2 + \alpha \gamma)y + \alpha \beta^2 z \equiv 0 \pmod{\alpha^3}.$

I will reveal the solution in due time (and it does not involve explicit computation of congruences), but if come up with a solution let me know!