Law of Quadratic Reciprocity

Theorem
Let $p$ and $q$ be distinct odd primes.

Then:
 * $\left({\dfrac p q}\right) \left({\dfrac q p}\right) = \left({-1}\right)^{\dfrac {\left({p - 1}\right) \left({q - 1}\right)} 4}$

where $\left({\dfrac p q}\right)$ and $\left({\dfrac q p}\right)$ are defined as the Legendre symbol.

An alternative formulation is: $\left({\dfrac p q}\right) = \begin{cases} \quad \left({\dfrac q p}\right) & : p \equiv 1 \lor q \equiv 1 \pmod 4 \\ -\left({\dfrac q p}\right) & : p \equiv q \equiv 3 \pmod 4 \end{cases}$

The fact that these formulations are equivalent is immediate.

This fact is known as the Law of Quadratic Reciprocity, or LQR for short.

Proof
Let $p$ and $q$ be distinct odd primes.

Consider the rectangle in the $x y$ plane with vertices at $\left({0, 0}\right), \left({\dfrac p 2, 0}\right), \left({\dfrac p 2, \dfrac q 2}\right), \left({0, \dfrac q 2}\right)$.

The number of lattice points inside this rectangle is $\dfrac {p - 1} 2 \times \dfrac {q - 1} 2$.


 * LQR.png

Consider the diagonal from $\left({0, 0}\right)$ to $\left({\dfrac p 2, \dfrac q 2}\right)$.

This has the equation:
 * $y = \dfrac q p x$

there were a lattice point $\left({a, b}\right)$ on the diagonal.

Then:
 * $b = \dfrac q p a$

But as $p$ and $q$ are distinct primes, this would mean:
 * $p$ divides $a$

and:
 * $q$ divides $b$

This means that $\left({a, b}\right)$ has to be outside the rectangle.

So there are no lattice points on the diagonal inside the rectangle.

Let $A$ and $B$ be the triangular regions inside the rectangle lying respectively above and below the diagonal.

Let $k \in \Z: 0 < k < \dfrac p 2$.

The number of lattice points in $B$ which lie directly above the point $\left({k, 0}\right)$ is:
 * $\left \lfloor {\dfrac q p k}\right \rfloor$

where $\left \lfloor {\dfrac q p k}\right \rfloor$ is the floor of $\dfrac q p k$.

So the total number of lattice points in $B$ is given by:
 * $\displaystyle N_B = \sum_{k \mathop = 1}^{\frac{p - 1} 2} \left \lfloor {\dfrac q p k}\right \rfloor$

Let $\alpha \left({q, p}\right)$ be defined as:
 * $\displaystyle \alpha \left({q, p}\right) = \sum_{k \mathop = 1}^{\frac{p-1}2} \left \lfloor {\dfrac q p k}\right \rfloor$

In the same way, by counting the lattice points to the right of $\left({0, k}\right)$, the total number of lattice points in $A$ is given by:
 * $N_A = \alpha \left({p, q}\right)$

Now the total number of lattice points in the rectangle is $N_A + N_B$.

But this is also equal to:
 * $\dfrac {p - 1} 2 \times \dfrac {q - 1} 2 = \dfrac {\left({p - 1}\right) \left({q - 1}\right)} 4$

So we have that:
 * $\alpha \left({p, q}\right) + \alpha \left({q, p}\right) = \dfrac {\left({p - 1}\right) \left({q - 1}\right)} 4$

The fact that these counts are equal relies upon the fact that there are no lattice points on the diagonal, as demonstrated above.

Now we invoke Eisenstein's Lemma:
 * $\left({\dfrac a p}\right) = \left({-1}\right)^{\alpha \left({a, p}\right)}$

So this gives us:

which is LQR.