# If $d$ divides $a^4+a^3+2a^2-4a+3$, prove that $d$ is a fourth power modulo $13$

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.

## 1 Answer

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 $$13$$.