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Let $\pi=a+bi\in \mathbb{Z}[i]$ be a Gaussian prime with $a$ and $b$ nonzero, and $b$ even. For odd rational primes $p=\pi\bar\pi$ and $q\neq p$, define $\pi^{\frac{1}{2}\left(q-\left(\frac{-1}{q}\right)\right)} = \alpha + \beta i$, then: \begin{align} \left(\frac{p}{q}\right)=-1 &\iff q\vert\alpha,\;\mathrm{and}\\ \left(\frac{p}{q}\right)=1 &\iff q\vert\beta. \end{align}

I was able to prove this using expansions of $\alpha\beta$ for $q=3\;\mathrm{mod}\;4$, and $p\alpha\beta$ for $q=1\;\mathrm{mod}\;4$, which is straightforward but not particularly insightful. I have searched the literature for a proof that ties this result to the broader concepts of quadratic residues, but I have not been able to find any mention of it.

Specifically, my question is, is this a known result? If so, is there a reference that ties it into the broader context of quadratic residues and reciprocity laws?

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1 Answer 1

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We have the following generalization: Let $K$ be a quadratic extension of $\mathbb Q$, $\mathcal O_K$ the ring of integers, $q$ an odd prime of $\mathbb Q$ unramified in $K$, and $x \in \mathcal O_K$ prime to $q$. Let $e = \frac{ q-1}{2}$ if $q$ splits in $K$ or $e=\frac{q+1}{2}$ if $q$ is inert in $K$.

Then $x^{e} \in \mathbb Z + q \mathcal O_K $ if and only if $\left( \frac{\operatorname{Norm}(x) }{q} \right)=1$ and $\operatorname{tr} (x^e) \in q \mathbb Z$ if and only if $\left( \frac{\operatorname{Norm}(x) }{q} \right)=-1$.

Proof: The key identity here is that $$x^q = \operatorname{Frob}_q(x) = \begin{cases} x & \textrm{if }q\textrm{ split} \\ \overline{x}& \textrm{if }q\textrm{ inert} \end{cases}$$ which combined with the usual formula for the quadratic residue gives (mod $q$)

$$\left( \frac{ \operatorname{Norm}(x)}{q} \right) = \left( \frac{ x\overline{x}}{q} \right) = x^{ \frac{q-1}{2}} {\overline{x}}^{\frac{q-1}{2}} = x^{ \frac{q-1}{2}} {\overline{x}}^{\frac{q-1}{2}} \cdot \frac{ \operatorname{Frob}_q(x)}{x^q} =\begin{cases} x^{ - \frac{q-1}{2}} \overline{x}^{\frac{q-1}{2}} & \textrm{if }q\textrm{ split} \\ x^{ - \frac{q+1}{2}} \overline{x}^{\frac{q+1}{2}} & \textrm{if }q\textrm{ inert} \end{cases}$$ $$ = \frac{ \overline{x}^e}{ x^e} = \frac{ \overline{x^e}}{x^e}.$$

If the quadratic residue symbol is $1$ this gives $\overline{x^e}=x^e$ (mod $q$), so $x^e \in \mathbb Z + q \mathcal O_K$, while if the quadratic residue symbol is $-1$ his gives $\overline{x^e}=-x^e$ (mod $q$), so $\operatorname{tr}(x^e) = x^e + \overline{x^e}=0$ mod $q$.

The proof should probably clarify a little more what's going on here.

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