# Describing the primes $p$ for which the Legendre symbol $(\frac{-6}{p})=1$

I would love your help with describing the primes $p$ for which the Legendre symbol $(\frac{-6}{p})=1$.

From the properties of the Legendre symbol I know that $(\frac{-6}{p})=(\frac{-1}{p})(\frac{2}{p})(\frac{3}{p})$,$(\frac{-1}{p})=(-1)^{\frac{p-1}{2}}$, and $(\frac{2}{p})=(-1)^{\frac{p^2-1}{8}}$. If I knew about a similar formula for $3$, it would help me, but I'm afraid there isn't one.

What should I do for solving this one?

Thanks!

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Actually, the main trick is that $$(-3 | p) = (p | 3)$$ and depends on the fact that $$3 \equiv 3 \pmod 4.$$ Quite a time saver. I use the horizontal typesetting of the symbol, which was introduced by L. E. Dickson. The vertical style always makes me think of fractions.

So, why? If $p \equiv 1 \pmod 4,$ then $$(-3|p) = (-1|p ) \cdot (3 |p) = 1 \cdot (p|3) = (p|3).$$

Switching to another letter, if prime $q \equiv 3 \pmod 4,$ then $$(-3|q) = (-1|q ) \cdot (3 |q) = -1 \cdot -(q|3) = (q|3).$$

What does a value of 1 tell us? if $(-24 | p) = (-6|p)=1$ for a prime $p \neq 2,3,$ we get either an expression $$p = u^2 + 6 v^2, \; \mbox{as in} \; \; \{ 7, 31, 73, 79, 97, 103, 127, \ldots \},$$ all of which are $\equiv 1 \; \mbox{or} \; 7 \pmod {24},$ or $$p = 2 x^2 + 3 y^2, \; \mbox{as in} \; \; \{ 5,11,29,53,59,83,101,107,131,149, \ldots \},$$ all of which are $\equiv 5 \; \mbox{or} \; 11 \pmod {24}.$ As I said, the primes $2,3$ are to be considered separately.

The same thing works with the Jacobi symbol, which is just a product of Legendre symbols. Just an example, $(-35 | p) = (p | 35).$ Note the required $35 \equiv 3 \pmod 4.$

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Is the «$3\equiv 3\pmod 4$» a typo? – Mariano Suárez-Alvarez May 23 '12 at 19:42
@Mariano, thank you for checking on things. No, it is a very difficult lemma from Tate's thesis. But think of the havoc if it had turned out that $3 \equiv 1 \pmod 4,$ we would never hear the end of it. – Will Jagy May 23 '12 at 19:55
:D ${}{}{}{}{}$ – Mariano Suárez-Alvarez May 23 '12 at 19:58
$(\frac{-6}{p})=(\frac{2}{p})(\frac{-3}{p})$ and when you use the chinese lemma you obtain $p\equiv 1,7,6,10 [24]$. Strange... – Maman Jun 9 at 16:06

Hint: Use the quadratic reciprocity law to express $\left(\frac{3}{p}\right)$ in terms of $\left( \frac{p}{3}\right)$.

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Thanks- O.k. so I know that if p is equivalent to$3 (mod 4)$ than $(\frac{3}{p})=-(\frac{p}{3})$ and if $p$ is equivalent to $1 (mod 3)$ so it's -1 and if $p$ is equivalent $2 (mod 3)$ so it's 1, what about all the other cases? – Jozef May 23 '12 at 15:42
If $p \equiv 1 \pmod 4$ then $\left( \frac{3}{p}\right) = \left(\frac{p}{3}\right)$, by quadratic reciprocity again. – marlu May 23 '12 at 15:49

Continuing with Marlu's answer and comment:

(1) $\,p=1\pmod 4\Longrightarrow \left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)=1\Longleftrightarrow p=0,1\pmod 3$

2) $\,p=3 \pmod 4\Longrightarrow \left(\frac{3}{p}\right)=-\left(\frac{p}{3}\right)=1\Longleftrightarrow p=2\pmod 3$

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