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Let $p$ denote a prime and let $f$ be the polynomial $x^3 - 7x - 7$, considered as an element of $\mathbb{F}_p[x]$. If $f$ is irreducible over $\mathbb{F}_p$, let $K$ denote a splitting field of $f$, and let $\alpha$ be any root of $f$ in $K$. A straightforward calculation shows that $3\alpha^2-5\alpha-14$ and $-3 \alpha^2+4\alpha+14$ are also roots of $f$, and that all three roots are distinct.

The Frobenius automorphism $\sigma$ of $K$ is defined by $\sigma(y) = y^p \text{ for all } y \in K$. The Frobenius automorphism permutes the roots of $f$ in $K$. Consequently, either $$\alpha^p = 3\alpha^2 - 5\alpha -14 $$ or $$\alpha^p = -3 \alpha^2 + 4 \alpha + 14$$

Empirically, I find that the first case holds when $p \equiv \pm 3 \pmod 7$ and that the second case holds when $p \equiv \pm 2 \pmod 7$. I have been unable to prove these assertions.

I'd like to have a proof of this or a pointer to relevant literature. For $n$ composite, I have investigated whether the analogous equations $\alpha^n = 3 \alpha^2 -5\alpha -14 \text{ in case } n\equiv \pm 3 \pmod 7$, $\alpha^n = -3 \alpha^2 + 4 \alpha + 14 \text{ in case } n \equiv \pm 2 \pmod 7$ can hold. Here, $\alpha$ is the equivalence class of $x$ in $\mathbb{Z}[x]/(f,n)$.

For $n$ up to $10^{11}$, I have found no solutions or pseudoprimes so far, although the search is only about half-complete.

David Bernier
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    Over $\Bbb{Q}$ the splitting field is cyclic given that the discriminant is a square ($49$). It sure looks like the splitting field should be the real subfield of $\Bbb{Q}(\zeta_7)$. If you can write the roots in terms of $\zeta_7$, such a relation (usually) survives over $\Bbb{F}_p$, and then you can settle the matter, because the action of the Frobenius of $\zeta_7$ is easy to work with. Indeed, the answer will depend on the residue class of $p$ modulo seven. – Jyrki Lahtonen May 24 '25 at 17:43
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    @JyrkiLahtonen Hi! Assuming this is the same field (over the rationals for me) as $x^3 + x^2 - 2x - 1,$ how might we (find out how to) map the roots of one to the roots of the other? The one I give is explicit cosines... Looking at your first comment again, maybe I know how to approach that, it has just been years.. Reuschle https://archive.org/details/tafelncomplexer00unkngoog/page/n21/mode/2up and chapter from Cox http://zakuski.math.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf – Will Jagy May 24 '25 at 18:13
  • Hi @WillJagy. In an earlier question the real roots of this polynomial were found to be $2\cos(4\pi/7)-2\cos(2\pi/7)$, $2\cos(8\pi/7)-2\cos(4\pi/7)$ and $2\cos(2\pi/7)-2\cos(8\pi/7)$. In other words, the quadratic $x^2-x-2$ needs to be applied to the roots of the cubic you gave. See the linked thread. – Jyrki Lahtonen May 24 '25 at 18:29
  • @JyrkiLahtonen Thank you – Will Jagy May 24 '25 at 18:40

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The hard work was carried out by Aphelli when they answered your previous question. Recycling the following from their answer.

The roots of $f(x)$ are $r_1=\zeta^2-\zeta-\zeta^{-1}+\zeta^{-2}$, $r_2=\zeta^4-\zeta^2-\zeta^{-2}+\zeta^{-4}$ and $r_3=\zeta^8-\zeta^4-\zeta^{-4}+\zeta^{-8}=-r_1-r_2$.

The upshot here is that $\zeta$ can be any primitive seventh root of unity in some extension field of $K$.

A straight forward calculation (we can use the real roots, or $\zeta=e^{2\pi i/7}$, to settle this!) shows that for the quadratic $q(x)=3x^2-5x-14$ we have $q(r_1)=r_3$. If $F:K\to K$ is the Frobenius automorphism of $K(\zeta)$, then $$ F(r_1)=\zeta^{2p}-\zeta^p-\zeta^{-p}+\zeta^{-2p}. $$ So when $p\equiv \pm 2\pmod 7$ we have $F(r_1)=r_2$, and when $p\equiv\pm3\pmod 7$, we have $F(r_1)=r_3$. This proves your observation.

Jyrki Lahtonen
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