Legendre symbols always match?

Question

When I test these Legendre symbols on the primes that are $p\equiv 1\bmod 12$, they always seem to equal each other: $$\left( \frac{-6+2\sqrt{-3}}{p}\right)=\left(\frac{3+2\sqrt{3}}{p}\right)$$ Is this true, and is there a way to show this?

Answer 1

It seems that $$\left( \frac{-6+2\sqrt{-3}}{p}\right)=\left(\frac{3+2\sqrt{3}}{p}\right) = \left (\frac{2\sqrt{3}}{p}\right)$$

I can prove the second equality. $$2\sqrt{3}(3 + 2\sqrt{3}) = 3(1 + \sqrt{3})^2$$ Hence, $$\left(\frac{3+2\sqrt{3}}{p}\right) = \left(\frac{2\sqrt{3}}{p}\right)$$ since $\left(\frac{3}{p}\right) = 1$.

Update:

Will Jagy noticed that the proof of $\left( \frac{-6+2\sqrt{-3}}{p}\right)= \left(\frac{2\sqrt{3}}{p}\right) $ is reduced to calculating $\left(\frac{2\sqrt{-1}}{p}\right)$. You can prove it like this: $$2\sqrt{3}(-6 + 2\sqrt{-3}) = -6\sqrt{-1}(1 + \sqrt{-3})^2$$ Hence, $$\left(\frac{2\sqrt{3} (-6+2\sqrt{-3})}{p}\right) =\left(\frac{-3}{p}\right)\left(\frac{2}{p}\right)\left(\frac{\sqrt{-1}}{p}\right) = 1 \cdot (-1)^{\frac{p^2 - 1 }{8}} \cdot (-1)^{\frac{p-1}{4}} = 1$$

Answer 2

There are some needed things: you have $p \equiv 1 \pmod 4.$ There is a split $\pmod 8:$

When $p \equiv 5 \pmod 8,$ $-1$ is not a fourth power $\pmod p,$ also $(2|p) = -1$

When $p \equiv 1 \pmod 8,$ $-1$ is a fourth power $\pmod p,$ also $(2|p) = 1$

Denis has shown that $\left(\frac{3+2\sqrt{3}}{p}\right) = \left(\frac{2\sqrt{3}}{p}\right)$

The other expression has: $$ - \sqrt{-3} (-6 + 2 \sqrt{-3}) = 6 + 6 \sqrt{-3} = (3 + \sqrt{-3})^2 $$ so that $\left(\frac{-6+2\sqrt{-3}}{p}\right) = \left(\frac{-\sqrt{-3}}{p}\right)$

The question of the OP amounts to the conjecture that

$$ ? \; \; \left(\frac{-\sqrt{-3} \cdot 2 \sqrt{3}}{p}\right) \; \; = \; \; 1 \; \; ?$$ or

$$ ? \; \;\left(\frac{-6\sqrt{-1} }{p}\right) \; \; = \; \; \left(\frac{2\sqrt{-1}}{p}\right) \; \; = \; \; 1 \; \; ?$$

From the beginning of this answer, either $2$ and $ \sqrt{-1}$ are both residues or both are nonresidues; in either case

$$ \left(\frac{2\sqrt{-1}}{p}\right) \; \; = \; \; 1$$

Short version: with $p \equiv 1 \pmod 4,$ we know that there is a $\sqrt{-1}.$ With that in hand we find

$$ (1 + \sqrt{-1})^2 = 2 \sqrt{-1} $$ so that, indeed,

$$ \color{red}{ \left(\frac{2\sqrt{-1}}{p}\right) \; \; = \; \; 1 }$$