How to show this Legendre symbol problem Let $n \in \mathbb{N}$ and $p$ an odd prime number such as $p \nmid n$.
Prove that:
$\exists x, y  \in \mathbb{Z};\,\,  \gcd(x, y) = 1$ such as $x^{2} + ny^{2} \equiv 0\,  (\mod p) \iff \Bigg(\displaystyle \frac{-n}{p}\Bigg) = 1$, where $\Bigg(\displaystyle \frac{-n}{p}\Bigg)$ represent the Legendre Symbol.
I've been trying but I have not idea where to start. Thanks in advance.

Here is an outline, see if you can fill in more details.
First part: if $x^2+ny^2\equiv0\pmod p$ then $y$ is not a multiple of $p$.  (If so then $x$ is a multiple of $p$, but this is impossible since $\gcd(x,y)=1$.)  So $y$ has a multiplicative inverse $z$ modulo $p$, so $-n\equiv(xz)^2\pmod p$.
Converse: if $\bigl(\frac{-n}p\bigr)=1$ then for some $x$ we have $-n\equiv x^2\pmod p$, that is, $x^2+n\equiv0\pmod p$, and now you should have no trouble finding a suitable value of $y$.
