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}


\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!



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