I was told an interesting fact that quadratic reciprocity follows from the modularity of the theta function $\theta(z) = \sum_{n \in \mathbb{Z}}e^{2\pi in^{2}z}$:
$$\theta(\gamma z) = \left(\frac{c}{d}\right)\epsilon_{d}^{-1}(cz+d)^{1/2}\theta(z)$$
where $\gamma = \begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in \Gamma_{0}(4)$, $\left(\frac{c}{d}\right)$ is the extended Jacobi symbol, and $\epsilon_{d}$ is the normalized Gauss sum attached to the quadratic character $(-/d)$. The idea is to set $z = -p/q+\epsilon$ and let $\epsilon > 0$. From this one get's the Landsberg-Schaar relation $$\frac{1}{\sqrt{p}}\sum_{n = 0}^{p-1}e^{\frac{2\pi in^{2}q}{p}} = \frac{e^{\pi i/4}}{\sqrt{2q}}\sum_{n = 0}^{2q-1}e^{-\frac{\pi in^{2}p}{2q}}.$$ When $q = 1$ we get Gauss's evaulation of the quadratic gauss sum and quadratic reciprocity can be proven from this (this was the original post this).
I'm curious about the converse question so to speak. That is, what can you prove about the theta function from quadratic reciprocity. I think it boils down to that the nebentypus
$$\left(\frac{c}{d}\right)\epsilon_{d}^{-1}$$
is a homomorphism. So if $\gamma' = \begin{pmatrix} a' & b' \\\ c' & d' \end{pmatrix}$ then this is equivalent to showing
$$\left(\frac{c'a+d'c}{c'b+d'd}\right)\epsilon_{d'd}^{-1} = \left(\frac{c'}{d'}\right)\left(\frac{c}{d}\right)\epsilon_{d'}^{-1}\epsilon_{d}^{-1}$$
because $c$ and $c'$ are equivalent to $0$ modulo $4$.
I've played around with this a lot and I can't seem to reduce it to quadratic reciprocity. Adding a multiple of the bottom of the symbol on the left to the top component doesnt do much because there is no information relating the primed letters to the nonprimed letters.