If $a$ is a quadratic residue of the prime $p= 8n+5,$ then the solutions of $x^2 \equiv a \ (\text{mod} \ p)$ are $x \equiv \pm a^{n+1}$ or $\pm 2^{2n+1}a^{n+1} \ (\text{mod} \ p)$
I have shown that $x \equiv \pm a^{n+1}$ or $\pm 2^{2n+1}a^{n+1} \ (\text{mod} \ p)$ are the possible solutions. How do I show they are all the possible solutions? Appreciate any advice, thank you.
$(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic of order $p-1$, generated by $b$ say. Let $a=b^n$ and $x=b^m$. Then $2m$ must agree with $n$, mod $p-1$. If $n$ is even then divide through by 2, so $m=n/2$ or $n/2+(p-1)/2$. If $n$ is odd then there is no solution. So either way there can't be more than two.