For which primes $p$ does $x^2\equiv3\pmod{p}$ have a solution?

Question

attempt at solution: using Legendre symbol, what I did was notice that the equation has a solution if ($3/p$)=1 IFF ($p/3$)x(-1)^(p-1)/2 = 1

So, can I say that p has to satisfy two properties (4 cases)? For example, case 1 is: If p≡1 mod3, then ($p/3$)=1, then we have a solution IFF p≡1 mod4

The statement "the equation has a solution if... iff..." is a little unclear. Can you please rephrase that? — barak manos, Feb 26 '15 at 17:23
You're on the right track. But you can put your conclusion in a nicer form by using the Chinese Remainder Theorem. — Brent Kerby, Feb 26 '15 at 17:26

score 0 · Answer 1 · edited Apr 13 '17 at 12:21

In this related question I proved that for every prime $p>3$, $-3$ is a quadratic residue iff $p\equiv 1\pmod{3}$ by using only the Cauchy theorem for groups. Since for every odd prime we know that $-1$ is a quadratic residue iff $p\equiv 1\pmod{4}$, and: $$\left(\frac{3}{p}\right)=\left(\frac{-3}{p}\right)\left(\frac{-1}{p}\right),$$ it follows that $3$ is a quadratic residue $\!\!\pmod{p}$ iff $\color{blue}{p\equiv\pm 1\pmod{12}}$.

For which primes $p$ does $x^2\equiv3\pmod{p}$ have a solution?

1 Answers1