Quadratic Gauss Sums to prove law of quadratic reciprocity

Question

Suppose we have $g_{2} = \sum_{t=0}^{p-1} \omega^{t^{2}}$, where $\omega$ denotes the p^th root of unity, and p is a prime equivalent to $1 \pmod{4}$. Let $q$ be a prime equivalent to $1 \pmod{4}$. How do I show that $g_{2} \in \mathbb{F}_{q} \iff $ $q$ is a quadratic residue mod ${p}$, without using the law of quadratic reciprocity. Similarly, how do I show that $p$ is a quadratic residue mod $q$ if and only if $q$ is a quadratic residue mod $p$, once again, without using the law of quadratic reciprocity?

Answer 1

Note that $g_2^2=\sum_{0 \leq k,l < p}{\omega^{k^2+l^2}}=\sum_{k \in \mathbb{F}_p}{c_k\omega^k}$, where $c_k=|\{(x,y) \in \mathbb{F}_p^2, x^2+y^2=k\}|$.

Now, let $i \in \mathbb{F}_p$ be a square root of $-1$. Then, $c_k=|\{(x,y) \in \mathbb{F}_p^2,\,(x+iy)(x-iy)=k\}|=|\{(x,y) \in \mathbb{F}_p^2,\,xy=k\}|$, so $c_k=p-1$ if $k \neq 0$ and $c_0=2p-1$. It follows $g_2^2=p$, thus $g_2 \in \mathbb{F}_q$ iff $p$ QR mod $q$.

Edit: I had misread the question, and indeed, I missed a whole part of it. If $q$ is a QR mod $p$, then the sequences $(t^2 \pmod{p})_{0 \leq t < p}$ and $(qt^2 \pmod{p})_{0 \leq t < p}$ are permutations one of the other. It follows $g_2^q=g_2$ hence $g_2 \in \mathbb{F}_q$.

However, we can see that if $q$ is not a QR mod $p$ and $g_2 \in \mathbb{F}_q$, then the reunion of two sequences is a permutation of $(t \pmod{p})_{0 \leq t < 2p}$, hence $g_2^q+g_2=0$, therefore $g_2=0$ and $p=g_2^2=0$, a contradiction.

So similarly $g_2 \in \mathbb{F}_q$ iff $q$ is a QR mod $p$.