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.
Best Answer
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 $p\equiv \pm 1\pmod{12}$.