With \(p=12k+1\), it's never the case that \(3\equiv0\mod p\), so certainly \(\left(\dfrac3p\right)
eq0\).
We can also eliminate the case that \(\left(\dfrac3p\right)=-1\) 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 \(\left(\dfrac3p\right)=1\).
