Solution
= Solution
For distinct odd primes $p,q$, <quadratic reciprocity> is
$$
\boxed{\left(\frac pq\right)\left(\frac qp\right)=(-1)^{(p-1)(q-1)/4}.}
$$
The usual supplementary laws are
$$
\left(\frac{-1}p\right)=(-1)^{(p-1)/2},\qquad
\left(\frac2p\right)=(-1)^{(p^2-1)/8}.
$$
The first supplement also follows immediately by inserting $a=-1$ into <Euler's criterion>.