The Quadratic Reciprocity Law

If \(p\) is an odd prime and \(a\) and \(b\) are relatively prime to \(p\), then $$ \begin{align} \left(\frac{a}{p}\right) &= \left(\frac{b}{p}\right), \quad \text{if} \quad a \equiv b \pmod{p} \\ \left(\frac{ab}{p}\right) &= \left(\frac{a}{p}\right) \left(\frac{b}{p}\right) \\ a^{\frac{p-1}{2}} &\equiv \left(\frac{a}{p}\right) \pmod{p} \end{align} $$

If \(p\) and \(q\) are distinct odd primes, then $$ \begin{align} \left(\frac{p}{q}\right) = \left(\frac{q}{p}\right) \end{align} $$ unless \(p \equiv q \equiv 3 \pmod{4}\) in which case, $$ \begin{align} \left(\frac{p}{q}\right) = -\left(\frac{q}{p}\right) \end{align} $$

If \(p\) is an odd prime, then $$ \begin{align} \left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}} = \begin{cases} 1, & \text{if } p \equiv 1 \pmod{4} \\ -1, & \text{if } p \equiv 3 \pmod{4} \end{cases} \end{align} $$ and $$ \begin{align} \left(\frac{2}{p}\right) = (-1)^{\frac{p^2-1}{8}} = \begin{cases} 1, & \text{if } p \equiv \pm 1 \pmod{8} \\ -1, & \text{if } p \equiv \pm 3 \pmod{8} \end{cases} \end{align} $$

If \(p\) is an odd prime and \(\gcd(a,p) = 1\), then the congruence $$ \begin{align} x^2 \equiv a \pmod{p^n} \end{align} $$ has a solution if and only if \(\left( \dfrac{a}{p} \right) = 1\)