Quadratic reciprocity (statement)

Theorem. For odd distinct primes $p,q$:
$$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{(p-1)(q-1)}{4}}.$$

Apply to $p=5$, $q=13$. $(5-1)/2=2$, $(13-1)/2=6$, so $(p-1)(q-1)/4=2\cdot6=12$, even. Thus $\left(\frac{5}{13}\right)=\left(\frac{13}{5}\right)=\left(\frac{3}{5}\right)$.

Compute. $3$ is a QNR mod $5$ (squares mod $5$: $1,4$), so $\left(\frac{3}{5}\right)=-1$.

Conclusion. $5$ is not a quadratic residue mod $13$.

Check. Squares mod $13$: $1,4,9,3,12,10$; $5$ not present.