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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!

3 Answers 3

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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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    $(\frac{-6}{p})=(\frac{2}{p})(\frac{-3}{p})$ and when you use the chinese lemma you obtain $p\equiv 1,7,6,10 [24]$. Strange...2016-06-09
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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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    If $p \equiv 1 \pmod 4$ then $\left( \frac{3}{p}\right) = \left(\frac{p}{3}\right)$, by quadratic reciprocity again.2012-05-23
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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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    @Ninja Google "Gauss quadratic reciprocity"2018-02-24