If $d$ divides $f(a)=a^4+a^3+2a^2-4a+3$, prove that $d$ is a fourth power modulo $13$.

$f(a)\equiv{(a-3)}^4\pmod {13}$. But how can we prove any divisor of $f(a)$ is a fourth power? If we prove that any prime divisor $p$ of $f(a)$ is a fourth power modulo $13$, we would be done as the fourth powers form a group under multiplication modulo $13$.

If we can write $f(a)=P(a)^2-13Q(a)^2$ for some polynomials $P(a)$ and $Q(a)$, $p$ divides $f(a)$ implies $13$ is a square modulo $p$ and quadratic reciprocity will imply $p$ is a square modulo $13$. I could not find suitable $P(a)$ and $Q(a)$ but I think this will help. To prove fourth power, I think quartic reciprocity will help, but I don't know.

Any help will be appreciated.


The key observation is that $K = \mathbf{Q}[x]/f(x)$ is not only Galois but is the degree $4$ subfield of $\mathbf{Q}(\zeta)$, where $\zeta$ is a 13th root of unity. Explicitly, the roots are $\zeta^n + \zeta^{3n} + \zeta^{9n}$ where $n = 1,2,4,8$. The Galois group is

$$G:=\mathrm{Gal}(K/\mathbf{Q})= (\mathbf{Z}/13 \mathbf{Z})^{\times}/\langle 3 \rangle \simeq \mathbf{Z}/4 \mathbf{Z},$$

where $[a] \in G$ is the element which sends $\zeta$ to $\zeta^a$. Note that the subgroup $\langle 3 \rangle$ is precisely the subgroup of $4$th powers modulo $13$. If $p$ divides $f(a)$, then $p$ splits completely in $K$ (as it is Galois). That means the Frobenius element at $p$ is trivial. But the Frobenius element sends $\zeta \mapsto \zeta^p$, so for this to be trivial $p \in \langle 3 \rangle$ or equivalently $p$ must be a $4$th power.

We see from this argument that the primes $p$ dividing $f(a)$ are exactly the primes which are $4$th powers modulo $13$.


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