I want to prove that $$\l(\frac 2p\r) = \begin{cases} 1 & p \equiv \pm 1 \pmod 8 \ -1 & \text{otherwise} \end{cases}.$$ I already have the proof for $\left(\frac{q}{p}\right) = \begin{cases} \left(\frac{p}{q}\right) & \text{if } p \text{ or } q \equiv 1 \pmod 4 \ -\left(\frac{p}{q}\right) & \text{if } p \equiv q \equiv 3 \pmod 4 \end{cases}$ and it says we can follow the steps of it?
#Proving quadratic reciprocity
mossy topaz
alpine kilnBOT
alpine kilnBOT
mossy topaz
Where for the first iff the following was used: \[5pt] Let $p$ be an odd prime, let $q\neq p$ be a prime, and let $d\mid p-1$. Then $q$ splits completely in $K_d$ if and only if $q$ is a $d$-th power modulo $p$, i.e.
[
q\equiv a^d \pmod p
]
for some $a\in\mathbb{Z}$.
And for the second: \[5pt] $n$ square-free, $p$ prime, $K=\mathbb{Q}(\sqrt{n})$.
(1) If $p\mid n$ then
[
p\mathcal{O}_K=(p,\sqrt{n})^2 \qquad\text{(ramified).}
]
(2) If $p=2$ then
[
2\mathcal{O}_K=
\begin{cases}
(2,1+\sqrt{n})^2, & \text{if } n\equiv 3\pmod 4 \quad\text{(ramified)},\[6pt]
\bigl(2,\tfrac{1+\sqrt{n}}{2}\bigr)\bigl(2,\tfrac{1-\sqrt{n}}{2}\bigr), & \text{if } n\equiv 1\pmod 8 \quad\text{(split)},\[6pt]
\text{prime (inert)}, & \text{if } n\equiv 5\pmod 8.
\end{cases}
]
(3) If $p\nmid 4n$ then
[
p\mathcal{O}_K=
\begin{cases}
(p,n+\sqrt{n})(p,n-\sqrt{n}), & \text{(split) if $m \equiv n^2 \pmod p$},\[6pt]
\text{prime (inert)}, & \text{if $m$ is not a square mod $p$}.
\end{cases}
]
mossy topaz
How can we "copy" this proof for 2/p now?
The hint goes "First prove that โ2 โ Q(ฯ8) and then proceed as in the proof of the other cases)."
I proved that. But how do we proceed?
alpine kilnBOT
@mossy topaz
alpine kilnBOT