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Problem 831

AIME late

Number theory Difficulty 5.6 Prove it

31.44. Let $p$ be a prime number. Prove that $\left(\frac{-3}{p}\right)=1$ for $p=6k+1$ and $\left(\frac{-3}{p}\right)=-1$ for $p=6k-1$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

31.44. It is clear that $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1)^{(p-1) / 2}\left(\frac{3}{p}\right)$ . Further, $(-1)^{(p-1) / 2}=1$ for $p=12 k+1$ and $p=12 k+5$ , and $(-1)^{(p-1) / 2}=-1$ for $p=12 k-1$ and $p=12 k-5$ . Using the result of problem 31.43, we obtain the required result.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.