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I’m having a really difficult time with the following proof involving the Legendre symbol:

Show that $\left(\dfrac{3}{p}\right) = 1$ iff $p \equiv \pm 1 \pmod{12}$

The normal tricks don’t seem to be working for this problem (ie this is one of a page of similar proofs). I’d appreciate any help offered in solving this problem.

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Do you know the quadratic reciprocity law? – Yoni Rozenshein Mar 27 '13 at 10:58
Combine information from $(-1)^{(p-1)/2}$ and $\left(\dfrac p3\right)?$ – Jyrki Lahtonen Mar 27 '13 at 11:05
Have you seen this? It directly solves your problem without the use of reciprocity, albeit the lemma in use could be used to derive the reciprocity. – awllower Mar 27 '13 at 11:20
up vote 3 down vote accepted

We use Reciprocity.

Suppose that $p$ is of the form $4k+1$. Then $(3/p)=(p/3)$. But $1$ is a quadratic residue of $3$ and $2$ is not. So in this case if $p\equiv 1\pmod{3}$, then $(3/p)=1$, and if $p\equiv 2\pmod{3}$, then $(3/p)=-1$.

If $p\equiv 3\pmod{4}$, then $(3/p)=-(p/3)$. We conclude as above that if $p\equiv 1\pmod{3}$ then $(3/p)=-1$, and if $p\equiv 2\pmod{3}$ then $(3/p)=1$.

The conditions modulo $3$ and $4$ can be combined to give conditions modulo $12$. For example, we got that $(3/p)=1$ when $p\equiv 1\pmod{4}$ and $p\equiv 1\pmod{3}$, and also when $p\equiv 3\pmod{4}$ and $p\equiv 2\pmod{3}$. Alternately, we can say that $(p/3)=1$ when $p\equiv \pm 1\pmod{12}$.

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