Does there exist a prime $p$ such that $X^4+X+1$ splits into a product of irreducible quadratics over $\mathbb F_p$?

Question

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.

Answer 1

$$ \huge 37 $$

Umm. The primes for which $x^4 + x + 1$ factors as two quadratics, together with the primes for which it factors as four linears, are those expressible with the indefinite quadratic form $$ p = x^2 + 15xy - y^2 $$ of discriminant $229.$

    37    53   173   193   229   241   347   359   383   439
   443   449   461   503   509   541   593   607   617   619
   643   691   907   967   977  1019  1051  1063  1097  1109
  1249  1277  1291  1303  1321  1399  1429  1583  1667  1741
  1783  1993  1997  2003  2087  2137  2143  2333  2347  2351
  2371  2381  2393  2503  2579  2657  2677  2687  2699  2729
  2749  2767  2791  2803  2897

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

The first three primes giving four linear factors are $193, 509, 593.$

jagy@phobeusjunior:~$ ./count_roots
193  count   1
509  count   2
593  count   3
1109  count   4
1291  count   5
1303  count   6
1399  count   7
1997  count   8
2087  count   9
2143  count   10
2371  count   11
2579  count   12
2677  count   13
2687  count   14
2791  count   15
3203  count   16
3761  count   17
3767  count   18
3881  count   19
3907  count   20
4019  count   21
4217  count   22
4397  count   23
4421  count   24
4547  count   25
4597  count   26
5119  count   27
5167  count   28
5209  count   29
5843  count   30
6007  count   31
6547  count   32
6701  count   33
7039  count   34
8123  count   35
8573  count   36
8669  count   37
8677  count   38
8713  count   39
8753  count   40
8849  count   41
8951  count   42
9049  count   43
9187  count   44
9403  count   45
9689  count   46
9851  count   47
9883  count   48
jagy@phobeusjunior:~$ 

Answer 2

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.

Answer 3

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.