Calculating $\left ( \frac{-3}{p} \right )$ when $p=u^2+3v^2$

I am trying to show that if $p=u^2+3v^2$ then we have that $\left ( \frac{-3}{p} \right )=1$ so that $p\equiv 1 \pmod 6$

Now my plan was to:

$$\left ( \frac{-3}{p} \right )=\left ( \frac{-1}{p} \right )\left ( \frac{3}{p} \right )$$

Then from quadratic reciprocity we have $\left ( \frac{3}{p} \right )\left ( \frac{p}{3} \right )=(-1)^\frac{p-1}{2}=1$ so that:

$\left ( \frac{3}{p} \right )=\left( \frac{p}{3} \right )$. We now have that as $2$ is primitive in $F_3$ that $p$ is a quadratic residue iff $p\equiv 1 \pmod 3$. Putting this together then gives that:

$$\left ( \frac{-3}{p} \right )=(-1)^\frac{p-1}{2}.\left ( \frac{3}{p} \right )=\left ( \frac{3}{p} \right ).$$

So I now need to show that $p\equiv 1 \pmod 3$. Now clearly $3$ does not divide $p$ (otherwise it would not be prime). Suppose that $p\equiv 2 \pmod 3\Rightarrow u^2\equiv 2 \pmod 3$ but this is not possible so we must have $p\equiv 1 \pmod 3$ which gives that $\left ( \frac{-3}{p} \right )=1$ as desired.

However in my solution is simply states that as $p=u^2+3v^2$ we have that $p$ does not divide $v$ and $(uv^{-1})^2\equiv -3 \pmod p$ so that $\left ( \frac{-3}{p} \right )=1$- could someone explain to me why this is true?

Thanks for any help

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Header gives wrong impression. –  ABC Apr 23 '13 at 12:10
@exploringnet why? –  hmmmm Apr 23 '13 at 12:39
@exploringnet How does the header if my question give the wrong impression exactly? –  hmmmm Apr 28 '13 at 20:28

1 Answer

We have $u^2 = -3v^2 \pmod p$

if $p|v$ then $p|u$ so $v$ is invertible mod $p$.

So -3 is a square mod p.

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