Prove that for prime $p > 3$ we have
$$ (\frac{3}{p}) = \begin{cases} (\frac{p}{3}) & \text{ if } p \equiv 1 \pmod {4} \\ -(\frac{p}{3}) & \text{ if } p \equiv 3 \pmod {4} \end{cases}$$
We start with the result from the previous exercise. For distinct odd primes $p$ and $q$,
$$ (\frac{p}{q}) \times (\frac{q}{p}) = \begin{cases} 1 & \text{ if } p \equiv 1 \pmod {4} \text{ or } q \equiv 1 \pmod {4} \\ -1 & \text{ if } p \equiv q \equiv 3 \pmod {4} \end{cases}$$
Setting $q=3$, requires $p>3$ since $p$ and $q$ are odd and distinct. This gives
$$ (\frac{p}{3}) \times (\frac{3}{p}) = \begin{cases} 1 & \text{ if } p \equiv 1 \pmod {4} \\ -1 & \text{ if } p \equiv 3 \pmod {4} \end{cases}$$
We consider each case:
- If $p \equiv 1 \pmod 4$, then $ (\frac{p}{3}) \times (\frac{3}{p}) =1$. Since the Legendre Symbols can only be 1 or -1, this is only true if both are 1 or -1. That is $ (\frac{p}{3})= (\frac{3}{p}) $
- If $p \equiv 3 \pmod 4$, then $ (\frac{p}{3}) \times (\frac{3}{p}) =-1$. Since the Legendre Symbols can only be 1 or -1, this is only true if one is -1 and the other 1. That is $ (\frac{p}{3})= -(\frac{3}{p}) $
We have shown that for prime $p > 3$,
$$ (\frac{3}{p}) = \begin{cases} (\frac{p}{3}) & \text{ if } p \equiv 1 \pmod {4} \\ -(\frac{p}{3}) & \text{ if } p \equiv 3 \pmod {4} \end{cases}$$