nt.number theory - Is there an approach to understanding solution counts to quadratic forms that doesn't involve modular forms?

The best short reference for this is Lehman Math Comp 1992 PDF. A more elaborate discussion is in ME and ALEX which appeared in the Journal of Number Theory, January 2012, volume 132, number 1, pages 258-275. I attempted to include this material in a final section of the JNT paper, that did not work out, see META

A (ternary) positive quadratic form is indicated by $langle a,b,c,r,s,t rangle$ which refers to
$$ f(x,y,z) = a x^2 + b y^2 + c z^2 + r y z + s z x + t x y. $$ The discriminant is
$$ Delta = 4 a b c + r s t - a r^2 - b s^2 - c t^2. $$
Forms are gathered together into genera when they are equivalent locally. A fundamental result of Siegel is that we may calculate the weighted average of representations, over a genus, of a given target number. Siegel's result relates quadratic forms and modular forms.

We have genera labelled $G_1, G_2, G_3.$ Given an odd prime $p,$ define useful integers $u,v$ such that
$$ (-u | p) = -1, ; ; ; (-v | p) = +1. $$
The first one, $G_1,$ is the only genus of discriminant $p^2.$ Then we have two of the six genera of discriminant $4 p^2,$ these have level $4 p$ and are classically integral. Together
$$
begin{array}{lccccc}
mbox{Genus} & Delta & mbox{Level} & mbox{2-adic} & mbox{$p$-adic} & mbox{Mass} \
G_1 & p^2 & 4 p & y z - x^2 & u x^2 + p(y^2 + u z^2) & (p-1)/48 \
G_2 & 4p^2 & 4 p & 2 y z - x^2 & u x^2 + p(y^2 + u z^2) & (p-1)/32 \
G_3 & 4p^2 & 4 p & x^2 + y^2 + z^2 & v x^2 + p(y^2 + v z^2) & (p+1)/96
end{array}
$$
Note that $G_1$ and $G_2$ represent exactly the same numbers, but with different representation measures. Furthermore, when $p equiv 3 pmod 4,$ then $h(x,y,z) = x^2 + p y^2 + p z^2 in G_2,$ but
when $p equiv 1 pmod 4,$ then $h(x,y,z) = x^2 + p y^2 + p z^2 in G_3.$

Let $s(n)$ be the number of representations of $n$ as the sum of three squares. Then, taking one form $g$ per equivalence class in the specified genus, let
$$ R_j(n) = sum_{g in G_j} frac{r_g(n)}{|mbox{Aut} g|} .$$

The two new identities are
$$ s(p^2 n) ; - ; p s(n) ; = ; 96 ; R_1(n); - ; 96 ; R_2(n), $$

$$ (p+2) ; s( n) ; - ; s(p^2 n) ; = ; 96 ; R_3(n) .$$