Show that if $p$ and $q$ are distinct odd primes then
$$ (\frac{p}{q}) \times (\frac{q}{p}) = \begin{cases} 1 & \text{ if } p \equiv 1 \pmod {4} \text{ or } q \equiv 1 \pmod {4} \\ -1 & \text{ if } p \equiv q \equiv 3 \pmod {4} \end{cases}$$
We start with the Law of Quadratic Reciprocity.
$$ (\frac{p}{q}) \times (\frac{q}{p}) = (-1)^{\frac{p-1}{2} \times \frac{q-1}{2}} $$
We write the distinct odd primes $p$ and $q$ as $p=4m + s$ and $4n + t$, where $m,n$ are integers and $s,t$ can be one of 1 or 3.
This means
$$ (\frac{p}{q}) \times (\frac{q}{p}) = (-1)^{(2m+\frac{s-1}{2}) \times(2n+\frac{t-1}{2})} $$
The index of $(-1)$ is
$$ (2m+\frac{s-1}{2}) \times(2n+\frac{t-1}{2}) $$
This index is even when either or both $s=1$ or $t=1$. This corresponds to $p \equiv 1 \pmod 4$ or $q \equiv 1 \pmod 4$. An even index means $ (\frac{p}{q}) \times (\frac{q}{p}) = 1$.
The index is odd when both $s=3$ and $t=3$. This corresponds to $p \equiv 3 \pmod 4$ and $q \equiv 3 \pmod 4$. An odd index means $ (\frac{p}{q}) \times (\frac{q}{p}) = -1$.
We have shown that if $p$ and $q$ are distinct odd primes then
$$ (\frac{p}{q}) \times (\frac{q}{p}) = \begin{cases} 1 & \text{ if } p \equiv 1 \pmod {4} \text{ or } q \equiv 1 \pmod {4} \\ -1 & \text{ if } p \equiv q \equiv 3 \pmod {4} \end{cases}$$