My professor worked this out in class and I am lost on how he did it. The definition I am using for split is, for a polynomial $f$ with degree d, with integer coefficients, a prime is $f$-split if $f\mod p$ has $d$ distinct roots.
What I want is to prove (understand the proof of) the set of primes that split $x^2-5$ is $\{p:p \equiv \pm1 \mod 5\}$.
A rough out line of the proof he did is, he stated there is a field $E$ that contains $F_{p}$ and $c \in E$ such that $\Phi_{5}(c) = 0$ Let $b = c+c^{-1}$. $\Phi_{5}(x)$ is the cyclotomic polynomial. with some manipulation he ended up with the desired congruency, yet I don't see the how he knew the choice of $b$ or even fully understand the proof. I added a photo of the given proof.
The assertion results, for odd primes, from the law of quadratic reciprocity, which says in this case that $5$ is a square $\bmod p$ if and only if $p$ is a square $\bmod 5$. Now the only non-zero squares $\bmod 5$ are $1$ and $-1$.