Does there exist a prime $p$ such that $X^4+X+1$ splits into a product of irreducible quadratics over $\mathbb F_p$? Does there exist a prime $p$ such that $X^4+X+1$ splits into a product of irreducible quadratics over $\mathbb F_p$?
I have checked a few primes but I just get a single linear factor out, or it is irreducible. This is just a curiosity of mine.
 A: As you've tagged this Galois theory, I assume this is an exercise relating to Dedekind's theorem. In view of this theorem, it suffices to show that the Galois group of $X^4+X+1$ over $\Bbb{Q}$ does not contain a product of two disjoint $2$-cycles, when identified with a subgroup of $S_4$.
As there are only three possible isomorphism types of Galois groups of polynomials of degree $4$, which are $S_4$, $A_4$ and $V_4$ (the Klein $4$-group), and each contains a product of two disjoint $2$-cycles, this is impossible. Even worse, Chebotarev's density theorem (in some sense a converse to Dedekind's theorem) tells you that there exist infinitely many primes for which the polynomial splits as a product of two irreducible quadratics.
A: Detailed hint (more like a proof strategy): 
The primes $p=2$ and $p=3$ are not good: over $\mathbb{F}_3$, $x=1$ is a root, and over $\mathbb{F}_2$, the only irreducible quadratic polynomial is $x^2+x+1$, whose square is not $x^4+x+1$. 
Hence, the characteristic of the field is different from $2$ and $3$. 
But then the Ferrari formula that solves the quadratic equation works: it only fails for fields of characteristic $2$ or $3$. 
Express the four roots, and try pairing the unary factors corresponding to them to form quadratic polynomials. 
You will obtain clear condition as for which numbers must be quadratic residues in the field in order for those polynomials to have coefficients in the prime field. 
Check if it is possible to satisfy one of these without having some root in the prime field. 
