For which $n$ do $A+A^n+A^{n^2} \equiv 0 \pmod n$, $A^{n^2+n+1} \equiv 7\mathrm{I} \pmod n$ hold, A the companion matrix of $x^3-7x-7$?

Question

Let $f=x^3-7x-7$. Suppose $p$ is a prime and $f$ is irreducible over $\mathbb{F}_p$, and let $K = \mathbb{F}_p[x]/(f)$, the field with $p^3$ elements. If we let $\alpha = x \pmod f$ then $\alpha$ is a root of $f$ in $K$. The map $\sigma: K \mapsto K$ defined by $\sigma(r)=r^p$ is an automorphism of $K$ (the Frobenius automorphism).

Because $\sigma$ is an automorphism of $K$ that fixes $\mathbb{F}_p$, a root of $f$ in $K$ is mapped by $\sigma$ to a root of $f$. Let $\beta:= \sigma(\alpha)$ and $\gamma:=\sigma(\beta)$ . The discriminant of $f$ over $\mathbb{Q}$ is $49$ which is non-zero modulo $p$ when $p$ is a prime other than $7$. Therefore $\alpha, \beta \text{ and } \gamma$ are the three distinct roots of $f$ in $K$.

The Galois group of $f$ over $\mathbb{F}_p$ is abelian, because $49$ is a perfect square. It follows that the trace of $\alpha$ is

$$\operatorname{Tr}_{K/\mathbb{F}_p} (\alpha) = \alpha + \beta + \gamma$$

and that the norm of $\alpha$ is

$$\operatorname{N}_{K/\mathbb{F}_p} (\alpha) = \alpha \beta \gamma.$$

Using Vieta’s formulas for the polynomial $f$, we get:

$$\alpha + \beta + \gamma = 0$$

$$\alpha \beta \gamma = 7 \pmod p.$$

Denoting by $A$ the companion matrix of $f \in \operatorname{GL}_{3}(\mathbb{F}_p)$, we get:

$$A + A^p + A^{p^2} = 0 \text{ in } \operatorname{Mat}_3(\mathbb{F}_p)$$ and $$A^{p^2+p+1} = 7\operatorname{I} \text{ in } \operatorname{Mat}_3(\mathbb{F}_p)$$

where $$ A = \begin{pmatrix} 0 & 0 & 7 \\ 1 & 0 & 7 \\ 0 & 1 & 0 \end{pmatrix} $$

(By abuse of notation, we continue below to use $A$ for the matrix in $\operatorname{GL}(n,\mathbb{Z})$)

My question is to characterize those natural numbers $n$ for which

$$A + A^n + A^{n^2} \equiv 0 \pmod n \tag 1$$

and separately

$$A^{n^2+n+1} \equiv 7\operatorname{I} \pmod n \tag 2$$

It has been observed empirically (for $n<10^9$) that if $n$ is a prime different from $7$, then each of congruences (1) or (2) is satisfied if and only if $n \equiv 2, 3, 4, \text { or } 5 \pmod 7$. Over the same range, the only composite satisfying (1) is $n = 4$, while there is no composite satisfying (2).

Answer 1

Please note: the bulk of this answer and its strongest result regard the case where $n$ is prime, where the OP’s conjecture is proved. The discussion of the composite case is more speculation and a few special cases.

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I claim the following: let $K$ be a field of characteristic not $7$ contained in some algebraic closed field $\overline{K}$.

If $u \in \overline{K}$ is a primitive $7$-th root of unity and $t_i=u^i+u^{-i}$ for $i=1,2,4$, then I claim that $x^3-7x-7=\prod_{i\in \{1,2,4\}}{(x-(t_{2i}-t_i)}$ where $t_8=t_1$.

Proof: every $t_i$ verifies $t_i^2-2=t_{2i}$ and $t_i^3+t_i^2-2t_i-1=0$ (because $\sum_{j=0}^6{u^j}=0$). A direct calculation shows that $t_{2i}-t_i=t_i^2-t_i-2$ is then always a root of $x^3-7x-7$.

Finally, the $t_{2i}-t_i$ are pairwise distinct because otherwise (after replacing $u$ with a power if needed) one has $t_4-t_2=t_2-t_1$, ie $u^4+u^{-4}-2u^2-2u^{-2}+u+u^{-1}=0$, which implies that $-3u^2-3u^{-2}-1=0$, so $3$ is invertible in $K$ and $t_2=-1/3$, whence $-1/27+1/9+2/3-1=0$ in $K$, whence $-1+3+18-27=0$ in $K$, thus $7=0$ in $K$, a contradiction.

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Now, we can define $t_i$ for any integer $i$ prime to $7$, and it is easy to see that $t_i$ only depends on the class of $\pm i \mod{7}$ (of which representatives are $1,2,4$).

In the algebraic closure of the field $\mathbb{F}_p$, the automorphism group is generated by the Frobenius $x \mapsto x^p$.

The action of Frobenius on $t_i$ is easily seen to be $t_i \mapsto t_{pi}$. Therefore, $t_{2i}-t_i$ is in $\mathbb{F}_p$ iff it is fixed by the Frobenius, iff $t_{2pi}-t_{pi}=t_{2i}-t_i$ iff (by the above argument) $\pm i \equiv \pm pi\mod{7}$ iff $p \equiv pm 1 \mod{7}$.

The upshot is that $f$ is reducible in $\mathbb{F}_p$ iff it has a root in $\mathbb{F}_p$ iff all its roots are in $\mathbb{F}_p$ iff $p \equiv \pm 1 \mod{7}$.

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In particular, your congruences (1) and (2) holds for every prime $n \equiv 2,3,4,5\mod{7}$.

As written in my comment, if $n \equiv 1\pmod{7}$ is prime, then $A$ is similar to a diagonal matrix in $\mathbb{F}_n$ (its characteristic polynomial has three distinct roots in $\mathbb{F}_n$), so $A^n=A \mod{n}$ and congruences (1) and (2) will never hold (for (2), note that $A^3=7A+7I$ has a seven as its $(2,1)$ coefficient).

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For $n$ composite, this gets (much) messier.

The point is that the roots of $f$ still exist as $\alpha, \beta, \gamma \in \mathbb{Z}[\zeta_7]/(n)$ (assume $n$ prime to $7$ to make matters simpler). Then (1) holds iff $\alpha+\alpha^n+\alpha^{n^2} \in n\mathbb{Z}[\zeta_7]$ (and same for $\beta,\gamma$).

Similarly, (2) holds iff $\alpha^{1+n+n^2}-7 \in n\mathbb{Z}[\zeta_7]$ (and same for $\beta,\gamma$).

Neither claim has a straightforward interpretation: one (obviously) wants to apply the CRT, but for instance for (2), the congruence class of $A^{1+n+n^2}$ mod some prime $p$ (say, if $p \mid n$) depends on the rsidue class of $n$ mod $p-1$ (if $p \equiv \pm 1 \mod{7}$) or $n$ mod $p^3-1$ otherwise, which is difficult to handle since we have to look at this simultaneously for every prime divisor of $n$.

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Here are some examples.

The roots of $f$ modulo $13$ are $1,2,10$, so $1$ is an eigenvalue of $A$ modulo $13$, from which is follows that there is no integer $n$ for which $A^{1+n+n^2} \equiv 7I \mod{13}$ or $A+A^n+A^{n^2}=0 \mod{13}$. In particular (1) and (2) never hold for multiples of $13$.

Another example: $9$ is an eigenvalue of $A$ modulo $659$, and $659$ is congruent to $1$ modulo $7$ and $-1$ modulo $4$, so $7$ is not a square modulo $659$, hence no power of $A$ can be congruent to $7I$ modulo $659$.

In particular (2) fails for every multiple of $659$.

(This generalizes to any divisor of an integer of the form $n^6-7n^2-7$ which is congruent to $1$ mod $7$ and to $—1$ modulo $4$ – of which there exists infinitely many, I think, by applying Cebotarev to the class of the Frobenius at $659$ in the Galois group over $\mathbb{Q}$ of the number field generated by the roots of $n^6-7n^2-7$ and $x^{28}-1$.)