Law of Quadratic Reciprocity Let p and q be odd primes. Then
\[ \left( \frac { p } { q } \right) \left( \frac { q } { p } \right) = ( - 1 ) ^ { \frac { p - 1 } { 2 } \frac { q - 1 } { 2 } } \iff \begin{cases} (\frac{p}{q})=(\frac{q}{p}) &\text{if either p or q is }1\bmod 4\\ (\frac{p}{q})=-(\frac{q}{p}) &\text{if both p and q are }3\bmod 4 \end{cases} \]
Proof. Note the ring isomorphism \( \mathbb{Z}_{pq}^{*} \cong \mathbb{Z}_{p}^* \times \mathbb{Z}_{q}^* \). Clearly they are also commutative ring, consider the quotient group \( G=\frac{\mathbb{Z}_{p}^* \times \mathbb{Z}_{q}^*}{ U } \) under multiplication, where \( U=\{ (1,1),(-1,-1) \} \). \( |G|=\frac{(p-1)(q-1)}{2} \)
Under \( \mathbb{Z}_{p}^* \times \mathbb{Z}_{q}^* \) :
\[ G=xU \text{ where } x \in X=\{ x_1 \times x_2 = (x_1,x_2) : 1\leq x_1 \leq p-1\ \land 1\leq x_2
\leq \frac{q-1}{2} \} \]
Under \( \mathbb{Z}_{pq}^{*} \) :
\begin{align*}
G &= y\{ 1, -1 \} \text{ where } y \in \{ (y,pq)=1 \land 1\leq y \leq \frac{pq-1}{2} \} \\
\iff G &= zU \text{ where } z\in Z=\{ (z_1,z_2) : z_1 \equiv y \bmod p \land z_2 \equiv y \bmod q \}
\end{align*}
\begin{align*}
\prod_{g\in G} &= ({(p-1)!}^{\frac{q-1}{2}}\ , \ {\frac{q-1}{2}!}^{(p-1)})U\\
&= ((-1)^{\frac{q-1}{2}}\ , \ {\frac{q-1}{2} !}^{2 \cdot \frac{p-1}{2}})U\\
&= ((-1)^{\frac{q-1}{2}}\ , \ (-1)^{\frac{p-1}{2}}\cdot (-1)^{\frac{p-1}{2}\cdot \frac{q-1}{2}})U\\
&= (\frac{(({pq-1})/{2})!}{\frac{p-1}{2}!\cdot \frac{q-1}{2}!\cdot p^\frac{q -1}{2}\cdot q^\frac{p-1}{2}}\ , \
\frac{(({pq-1})/{2})!}{\frac{p-1}{2}!\cdot \frac{q-1}{2}!\cdot p^\frac{q -1}{2}\cdot q^\frac{p-1}{2}})U
\end{align*}
Note:
\begin{align*}
&\quad \frac{(({pq-1})/{2})!}{\frac{p-1}{2}!\cdot \frac{q-1}{2}!\cdot p^\frac{q -1}{2}\cdot q^\frac{p-1}{2}}\\
&\equiv \frac{\left( \prod _ { i = 1 } ^ { p - 1 } i \right) \left( \prod _ { i = 1 } ^ { p - 1 } p + i \right)
\cdots \left( \prod _ { i = 1 } ^ { p - 1 } \left( \frac { q - 1 } { 2 } - 1 \right) p + i \right) \left( \prod _ {
i = 1 } ^ { \frac { p - 1 } { 2 } } \frac { q - 1 } { 2 } p + i \right)}{\frac{p-1}{2}!\cdot q^\frac{p-1}{2}}\\
&\equiv \frac{(-1)^{\frac{q-1}{2}}\cdot \frac{p-1}{2}!}{\frac{p-1}{2}!\cdot q^\frac{p-1}{2}}\\
&\equiv (-1)^{\frac{q-1}{2}}\cdot q^\frac{p-1}{2} \\
&\equiv (-1)^{\frac{q-1}{2}} \cdot (\frac{q}{p}) \mod p
\end{align*}
\[
\implies \prod_{g\in G}=((-1)^{\frac{q-1}{2}} \cdot (\frac{q}{p})\ ,\ (-1)^{\frac{p-1}{2}} \cdot (\frac{p}{q}))U
\]
\( (a_1,a_2)U=(a_3,a_4)U \implies a_1a_2 \equiv a_3a_4 \mod pq \)
\begin{align*}
&\implies (-1)^{\frac{p-1}{2}} \cdot(-1)^{\frac{q-1}{2}} \cdot \left( \frac { p } { q } \right) \left( \frac { q
} { p } \right) \equiv (-1)^{\frac{p-1}{2}} \cdot(-1)^{\frac{q-1}{2}} \cdot ( - 1 ) ^ { \frac { p - 1 } { 2 } \frac
{ q - 1 } { 2 } } \mod pq\\
&\implies \left( \frac { p } { q } \right) \left( \frac { q } { p } \right) = ( - 1 ) ^ { \frac { p - 1 } { 2 }
\frac { q - 1 } { 2 } }
\end{align*}
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PREVIOUSGauss's lemma