show that there exists and integer $x$ such that $x^2 \equiv -7 \ mod \ p$

Question

Let $\tau = ({1/2})(-1 + \sqrt{(-7)})$ and $R = \{a + b\tau \ | \ a,b \in Z\}$. The first part of the question is show $R's$ a ring and show that $N(a+b\tau) = a^2 - ab + 2b^2$ which is all easy ok then we suppose $p$ is a prime number such that $p$ is not irreducible in $R$, and we're asked to show that $p=\pi \pi *$ such that $N(\pi) = p$, thats the same as when you show this result for the guassian integers. If $\pi = a+b\tau$ then $$a^2 - ab + 2b^2 = p$$ But then it asks to show that there exists and integer $x$ such that $$x^2 \equiv -7 \ mod \ p$$ and I don't see how to do this. After this it asks for a sort of converse, if p is a prime number and $-7$ is a quadratic residue modulo p, prove that $R$ contains at least one, and at most four elements with norm p.

Answer 1

If $Q$ is a prime ideal in $R=\mathcal{O}_{\sqrt{-7}}$ with $N(Q)=p$, then we can write $Q=(p,a+\frac{1+\sqrt{-7}}{2})$ for some $a\in \mathbb{Z}$. Hence $$ p\mid N\left(a+\frac{1+\sqrt{-7}}{2}\right)=\left(a+\frac{1+\sqrt{-7}}{2}\right)\left(a+\frac{1-\sqrt{-7}}{2}\right)=((2a+1)^2+7)/4. $$ Hence we have $x^2+7\equiv 0\bmod p$ for $x=2a+1$.