# Second Supplement to Law of Quadratic Reciprocity

## Theorem

$\paren {\dfrac 2 p} = \paren {-1}^{\paren {p^2 - 1} / 8} = \begin{cases} +1 & : p \equiv \pm 1 \pmod 8 \\ -1 & : p \equiv \pm 3 \pmod 8 \end{cases}$

where $\paren {\dfrac 2 p}$ is defined as the Legendre symbol.

## Proof

Consider the numbers in the set $S = \set {2 \times 1, 2 \times 2, 2 \times 3, \dots, 2 \times \dfrac {p - 1} 2} = \set {2, 4, 6, \dots, p - 1}$.

From Gauss's Lemma:

$\paren {\dfrac 2 p} = \paren {-1}^n$

where $n$ is the number of elements in $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$.

As they are, the elements of $S$ are already least positive residues of $p$ (as they are all less than $p$).

What we need to do is count how many are greater than $\dfrac p 2$.

We see that:

$2 k > \dfrac p 2 \iff k > \dfrac p 4$

So the first $\floor {\dfrac p 4}$ elements of $S$ are not greater than $\dfrac p 2$, where $\floor {\dfrac p 4}$ is the floor function of $\dfrac p 4$.

The rest of the elements of $S$ are greater than $\dfrac p 2$.

So we have:

$n = \dfrac {p - 1} 2 - \floor {\dfrac p 4}$

Consider the four possible residue classes modulo $8$ of the odd prime $p$.

$p = 8 k + 1$:

 $\ds p$ $=$ $\ds 8 k + 1$ $\ds \leadsto \ \$ $\ds n$ $=$ $\ds 4 k - \floor {2 k + \frac 1 4}$ $\ds$ $=$ $\ds 4 k - 2 k$ $\ds$ $=$ $\ds 2k$

$p = 8 k + 3$:

 $\ds p$ $=$ $\ds 8 k + 3$ $\ds \leadsto \ \$ $\ds n$ $=$ $\ds 4 k + 1 - \floor {2 k + \frac 3 4}$ $\ds$ $=$ $\ds 4 k + 1 - 2 k$ $\ds$ $=$ $\ds 2 k + 1$

$p = 8 k + 5$:

 $\ds p$ $=$ $\ds 8 k + 5$ $\ds \leadsto \ \$ $\ds n$ $=$ $\ds 4 k + 2 - \floor {2 k + \frac 5 4}$ $\ds$ $=$ $\ds 4 k + 2 - \paren {2 k + 1}$ $\ds$ $=$ $\ds 2 k + 1$

$p = 8 k + 7$:

 $\ds p$ $=$ $\ds 8 k + 7$ $\ds \leadsto \ \$ $\ds n$ $=$ $\ds 4 k + 3 - \floor {2 k + \frac 7 4}$ $\ds$ $=$ $\ds 4 k + 3 - \paren {2 k + 1}$ $\ds$ $=$ $\ds 2 k + 2$

We see that $n$ is even when $p = 8 k + 1$ or $p = 8 k + 7$ and odd in the other two cases.

So from Gauss's Lemma, we have:

 $\ds \paren {\dfrac 2 p}$ $=$ $\ds \paren {-1}^n = 1$ when $p = 8 k + 1$ or $p = 8 k + 7$ $\ds \paren {\dfrac 2 p}$ $=$ $\ds \paren {-1}^n = -1$ when $p = 8 k + 3$ or $p = 8 k + 5$

As $7 \equiv -1 \pmod 8$ and $5 \equiv -3 \pmod 8$ the result follows.

$\blacksquare$