Question

Show that if p=12k+1 for some k, then (3/p)=1 where (3/p) is the legendre symbol

Answer 1

With
`p=12k+1`, it's never the case that
`3\equiv0\mod p`, so certainly
`\left(\frac3p\right)\\eq0`.

We can also eliminate the case that
`\left(\frac3p\right)=-1` by coming up with a counter-example. Notice that
`p=13` for
`k=1`, and that
`16=4^2\equiv3\mod13`.

So because there exists some
`x` such that
`x^2\equiv3\mod p` it follows that
`\left(\frac3p\right)=1`.