Simple comparison of positive ternary quadratic form representation counts

Question

Something came up yesterday in a referee request and I was surprised to find that I did not know the facts in full generality. This is about positive quadratic forms in three variables with integer coefficients, primitive, meaning the GCD of coefficients is 1.

In the case of the sum of three squares, if $R(n)$ is the total number of integral representations of $n,$ and $p$ an odd prime, it follows (induction on the highest power of $p$ dividing $n$) from the result in Hirschhorn and Sellers that $$ R(n p^2) \geq p R(n). $$

The relation is $$ R(n p^2) = (p+1 - (-n|p)) R(n) - p R(n/p^2), $$ where $(-n|p) $ is taken as $0$ when $p$ divides $n,$ and $R(n/p^2)$ is taken as $0$ when $p^2$ does not divide $n.$ See Is there a simple way to compute the number of ways to write a positive integer as the sum of three squares?

On the other hand, $$ R(4n) = R(n). $$

So, very simple from a modular forms perspective, I imagine, given some other positive ternary form $f,$ representation count function $R(n),$ and a prime $p$ (possibly $2$) for which $f(x,y,z)$ is isotropic in $\mathbb Q_p,$ is it true that

$$ R(n p^2) \geq p R(n)? $$

Note: I've been running experiments, it seems to be alright if $p$ divides the discriminant of $f,$ as long as isotropic as I said. I suspect this is trivial, with key words such as Eisenstein series, Hecke operators, other things the kids enjoy.

Answer 1

No, this isn't true in general. You could, for example, take $f(x,y,z) = x^{2} + y^{2} + 13z^{2}$. Then $R(1) = 4$ and $R(25) = 12$. The form is isotropic at $5$ (since $5$ doesn't divide the discriminant). Why might you get the idea that it would be true based on the sums of three squares form? The reason is that the theta series for the sums of three squares form is an eigenfunction for all the Hecke operators (which is encoded by the relations $R(np^{2}) = (p+1 - \left(\frac{-n}{p}\right)) R(n) - p R(n/p^{2})$ and $R(4n) = R(n)$), while in general, the theta series for other ternary forms will not be. It should be easy to show that if $f(x,y,z)$ is such that its theta series is a Hecke eigenform, then $f(x,y,z)$ must be regular.