Evaluating $\left(\frac{-2}{p}\right)$

Question

Prove that for $p\ge 3$, a prime: $$\left(\frac{-2}{p}\right) = \begin{cases} 1 & p\equiv1,3\pmod{8}\\ -1 & p\equiv-1,-3\pmod{8} \end{cases}$$

I already now that:

$$\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p^{2}-1}{8}}=(-1)^{\frac{(p-1)(p^{2}-1)}{8}}=(-1)^{\frac{p-1}{2}\cdot\frac{p+1}{2}\cdot\frac{p-1}{4}}$$

And I'm trying to show when the exponent is even (for the $1$ case).

We have:

1. $\frac{p-1}{4} = 2k \implies p- 1 = 8k \implies p = 8k + 1$. So we can infer that if $p\equiv 1\pmod{8}$ then $\left( \frac{-2}{p} \right)=1$
2. $\frac{p-1}{2}$ - doesn't yield anything new to us (I think)
3. $\frac{p+1}{2} = 2t \implies p+1 = 4t \implies p = 4t-1$

but I don't really see how to infer the other option $(p \equiv 3 \pmod{8})$

Answer 1

There's an error in your computation: $$\binom{-2}p=(-1)^{\tfrac{p-1}{2}}(-1)^{\tfrac{p^{2}-1}{8}}=(-1)^{\tfrac{(p-1)\color{red}{+}(p^{2}-1)}{8}}=(-1)^{\tfrac{p-1}{2}\bigl(1+\tfrac{p+1}{4}\bigr)}=(-1)^{\tfrac{p-1}{2}\tfrac{p+5}{4}}$$

This is equal to $1$, i.e. $-2$ is a square mod. $p$, if and only if $\;\dfrac12\dfrac{p-1}2\,\dfrac{p+5}2\,$ is even. Observe that $$\dfrac{p-1}2\;\text{is}\:\begin{cases}\text{even}\\\text{odd}\end{cases}\iff \dfrac{p+5}2\;\text{is}\:\begin{cases}\text{odd}\\\text{even}\end{cases}$$ so we have the following cases:

• if $\dfrac{p-1}2$ is even, we have $\;\dbinom{-2}p=1\iff\dfrac{p-1}2\equiv 0\mod 4\iff p\equiv 1\mod 8$,
• if $\dfrac{p-1}2$ is odd, $\;\dbinom{-2}p=1\iff\dfrac{p+5}2\equiv 0\mod 4\iff p\equiv -5\equiv 3\mod 8$.

Answer 2

If $p\equiv 3\pmod{8}$ then $\frac{p-1}{2}$ is even and $\frac{p^2-1}{8}$ is odd, so $\left(\frac{-2}{p}\right)=-1$.

Here you may find a proof avoiding quadratic reciprocity and exploiting some field theory.

Answer 3

$p=4t-1=4(t-1)+3$, so $p=3$ mod 4. Modulo 8 there are two possibilities:

1) $p=3$ (mod 8)

2) $p=7$ (mod 8)

By plugging in the numbers, 2) doesn't work, and 1) work

Edit:

There seems to be a mistake, remember that $(-1)^\frac{p-1}{2}(-1)^\frac{p^2-1}{8}=(-1)^{\frac{p-1}{2}+\frac{p^2-1}{8}}=(-1)^\frac{(p-1)(p+5)}{8}$