Squares in a finite field $\mathbb{F}_p$ I need to find all prime $p$ s.t $n+3$ is the inverse of $n-3$ in $\mathbb{F}_p$. So obviously this means $(n+3)(n-3)=1\mod p$, meaning $n^2=10\mod p$. So the question is - for which $p$ does the equation $x^2-10=0$ has a solution in $\mathbb{F}_p$, but I'm not quite sure how to approach this.
I tried using $(\frac{10}{p})=10^{\frac{p-1}{2}}\mod p$ but this got me nowhere. Any help would be apperciated.
 A: I assume the problem is to determine all primes for which there exists an $n$ for which $n+3$ is the inverse of $n-3$ in $\mathbb{F}_p$. As you note, this requires finding an $n$ such that $n^2\equiv 10\pmod{p}$, i.e., find the primes for which $10$ is a quadratic residue. Which of course just begs for Quadratic Reciprocity.
If $p=2$, then any even $n$ will work. If $p=5$, then any multiple of $5$ will do. So we may assume that $p\neq 2,5$.
We know that, from the quadratic character of $2$, that
$$\left(\frac{2}{p}\right) = \left\{\begin{array}{ll}
1 &\text{if }p\equiv \pm1 \pmod{8}\\
-1 &\text{if }p\equiv \pm3 \pmod{8}.
\end{array}\right.$$
And from Quadratic Reciprocity, we have
$$\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right)(-1)^{\frac{p-1}{2}\,\frac{5-1}{2}} = \left(\frac{p}{5}\right) = \left\{\begin{array}{ll}
1 & \text{if }p\equiv \pm 1\pmod{5},\\
-1 &\text{if }p\equiv \pm 3\pmod{5}.
\end{array}\right.$$
Since $\left(\frac{10}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{5}{p}\right)$, we see that $10$ is a square modulo $p$ if and only if both $2$ and $5$ are squares modulo $p$, or neither is. So there are $8$ remainders modulo $40$ where it is a square (four for the possibilities that it be $\pm 1$ modulo $8$ and $\pm 1$ modulo $5$; and four for when it is $\pm 3$ modulo $8$ and $\pm 3$ modulo $5$), and $8$ remainders modulo $40$ where it is not a square (the remaining combinations/remainders modulo $40$ that are relatively prime to $40$).  So use the Chinese Remainder Theorem to figure out those values.
