The following is a really wonderful theorem that I really have no idea how to prove.
Consider $p=ax^2+bxy+cy^2$, and let $\text{SL}_2\mathbb{Z}$ act on all such $p$ by $\begin{pmatrix} a&b \\ c&d \end{pmatrix}:x^i y^j \mapsto(ax+cy)^i (bx+dy)^j$ extended linearly.
Let $f(a(p),b(p),c(p))$ be a polynomial, where $a(p)$ is the coefficient of $x^2$ in $p$. $\text{SL}_2\mathbb{Z}$ acts on all such polynomials in $a,b,c$, call these polynomials $P$, by an extension of its action on $p$.
Prove that if $h \in P$ is fixed by all of $\text{SL}_2\mathbb{Z}$ then $h$ is a polynomial in $\text{disc}(p)=b^2-4ac.$
(See about 3:30 here).