Prime divisors of $5a^4-5a^2+1$

The following is an attempt to reverse engineer my comment under the OP to a solution requiring no algebraic number theory. To an extent I replace that with basic facts about finite fields. I think a solution with even more elementary tools is out there.

Assume that $p$ is a prime factor of $f(a)=5a^4-5a^2+1$ for some integer $a$. Clearly $f(a)$ is odd and congruent to $1$ modulo five, so $p\neq2,5$. This implies that any eventual zeros of the twentieth cyclotomic polynomial $$ \Phi_{20}(x)=x^8-x^6+x^4-x^2+1 $$ in some field $K$ of characteristic $p$ have multiplicative order twenty (the zeros of $h(x)=x^{20}-1$ in its splitting field over $K$ are all simple, because $h'(x)=20x^{19}$ is coprime to $h(x)$).

Consider the equation $$ \frac1a=z+\frac1z\Longleftrightarrow z^2-a^{-1}z+1=0\qquad(*) $$ over the field $\Bbb{F}_p$. This is quadratic in $z$, so it has a solution either in the field $\Bbb{F}_p$ itself or in its quadratic extension $\Bbb{F}_{p^2}$. We shall treat these two cases separately, but we first make the observations that $$ (z+\frac1z)^4-5(z+\frac1z)^2+5=\frac{\Phi_{20}(z)}{z^4}, $$ and also that (as a consequence of $(*)$) $$ (z+\frac1z)^4-5(z+\frac1z)^2+5=\frac{1-5a^2+5a^4}{a^4}=\frac{f(a)}{a^4}=0. $$ Therefore we can conclude that $z$ has multiplicative order twenty.

1. If $z\in\Bbb{F}_p^*$ this immediately implies that $20\mid p-1$. This is because the group $\Bbb{F}_p^*$ is cyclic of order $p-1$. In this case we must have $p\equiv1\pmod{20}$.
2. If $z$ is an element of the quadratic extension of $\Bbb{F}_p$ instead, then the polynomial $g(x)=x^2-a^{-1}x+1$ is irreducible in $\Bbb{F}_p[x]$. But, from $(*)$ we see that the zeros of $g(x)$ are $z$ and $1/z$. By Galois theory they are also $\Bbb{F}_p$-conjugates, so we can deduce that $1/z=z^p$. Therefore we also have that $z^{p+1}=1$. Consequently $20\mid p+1$, and in this case we have $p\equiv-1\pmod{20}.$