How does one compute the value of $\alpha^{p+2} + \beta^{p+2} + \gamma^{p+2}$ in $F_{p}$?

Question

Let $f = X^3 - 3X - 1$ and let $p$ be a prime such that $f$ is irreducible over $F_{p}$. Let $L$ be a splitting field of $f$ over $F_{p}$. For $k \geqslant 0$, let $V_{k} = \alpha^k + \beta^k + \gamma^k$ , where $\alpha$, $\beta$ and $\gamma$ are the three roots of $f$ over $L$. Let $\sigma$ be the Frobenius automorphism of $L$ defined by $\sigma(x)=x^p$. According to an online reference, the Galois group $Gal(L/F_{p})$ is cyclical and is generated by the Frobenius automorphism. Without loss of generality, suppose $\sigma(\alpha)=\beta$, $\sigma(\beta) = \gamma$, $\sigma(\gamma)=\alpha$. I was able to determine $V_{p}$ and $V_{p+1}$ , but I had trouble with $V_{p+2}$. $$V_{p+2} = \alpha^p \alpha^2 + \beta^p \beta^2 + \gamma^p \gamma^2 = \beta \alpha^2 + \gamma \beta^2 + \alpha \gamma^2.$$ It is easy to show that $V_{p+2}$ remains fixed under the action of $\sigma$. Numerical experimentation suggests that $V_{p+2} == +3$ or $V_{p+2}== -6$. Is this true ? If so how can one prove it?

Answer 1

Assume for now that $p \neq 2$. Let $S=\beta\alpha^2+\gamma\beta^2+\alpha\gamma^2$ and $T=\gamma\alpha^2+\alpha\beta^2+\beta\gamma^2$.

Then you can check that $S+T=V_1V_2-V_3=0-3V_1-3=-3$.

Now, $$ST=\beta\gamma\alpha(\alpha^3+\beta^3+\gamma^3)+\sum_{cyc}{\alpha^3\beta^3}+3\alpha^2\beta^2\gamma^2=1\cdot 3+3+\sum_{cyc}{\alpha^3\beta^3}=6+\frac{1}{2}(V_3^2-V_6).$$

Now, $V_6=3V_4+V_3=3(3V_2+V_1)+3=9V_2+3=3+9(V_1^2-2\sum_{cyc}{\alpha\beta})=3+9(0-2\cdot (-3))=57$.

Hence $ST=6+\frac{9-57}{2}=6-24=-18$.

Thus $S^2+3S-18=0$, whence $S=3$ or $S=-6$.

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Now, with a lucky guess, we can actually determine $V_{p+2}$. We allow $p=2$ in what follows.

If we work in a sufficiently large extension of $\mathbb{F}_p$, we can write $\alpha=u+u^{-1}$. Then $(u+u^{-1})^3=3u+3u^{-1}+1$, so $u^3-1+u^{-3}$, whence $u^3$ is a primitive $6$-th root of unity and $-u$ is a primitive $9$-th root of unity.

So we can write $\alpha=-(u+u^{—1}), \beta=-(v+v^{-1}), \gamma=-(w+w^{-1}$, where $\{u,v,w,u^{-1},v^{-1},w^{-1}\}$ are the primitive $9$-th roots of unity in $\overline{\mathbb{F}_p}$.

Let $S_t$ be the sum of the $x^t$, where $x$ runs through the primitive $9$-th roots of unity in $\overline{\mathbb{F}_p}$.

Clearly, $S_t$ only depends on the gcd of $t$ with $9$, and $S_1=0$ (consider the cyclotomic polynomial $\Phi_9$), $S_3=-3$ (every primitive $3$-rd root of unity appears three times and consider the cyclotomic polynomial $\Phi_3$) and $S_9=6$.

Thus $$-V_{p+2}=(u+u^{-1})^2(u^p+u^{-p})+(v^{-1}+v)^2(v^p+v^{-p})+(w+w^{-1})^2(w+w^{-p}) = S_{p+2}+2S_p+S_{p-2}=S_{p+2}+S_{p-2}.$$

Note that if $f$ is irreducible over $\mathbb{F}_p$, then $\{u,u^{-1}\} \neq \{u^p,u^{-p}\}$ whence $p\not\equiv \pm 1\pmod{9}$. Note also that $p \neq 3$. Hence exactly one number (let’s call it $k$) among $p\pm 2$ is divisible by $3$ and $V_{p+2}=-S_k$.

If $p \not\equiv \pm 2 \pmod{9}$, then $k$ is not divisible by $9$ so $V_{p+2}=-S_3=3$.

If $p \equiv \pm 2\pmod{9}$, then $k$ is divisible by $9$ so $V_{p+2}=-S_6=-6$.

Answer 2

$\newcommand{\aa}{\alpha}\newcommand{\bb}{\beta}\newcommand{\cc}{\gamma}$Let us first assume that $p \neq 2$.

Let $e_{1} = \aa + \bb + \cc$, $e_2 = \aa \bb + \aa \cc + \bb \cc$, $e_3 = \aa \bb \cc$, and $\Delta = (\aa - \bb)(\bb-\cc)(\cc-\aa)$.
Then, for $p \neq 2$, one verifies that $$\aa^{2} \bb + \bb^{2} \cc + \cc^{2} \aa = \frac{1}{2} e_{1} e_{2} - \frac{3}{2} e_{3} - \frac{1}{2} \Delta.$$ (The above is an identity that holds in any field of characteristic not two. Verifying it is not difficult: just expand the right-hand-side.)

Now, in our example, we have $e_{1} = 0$ and $e_{3} = 1$. So, by your observation, we get $$V_{p+2} = -\frac{3}{2} -\frac{1}{2} \Delta.$$

Now, it is a question of computing $\Delta$. The square of $\Delta$ is the well-known quantity called the discriminant. For the cubic at hand, we get $$\Delta^{2} = -4(-27) - 27(1) = 81.$$ (For example, the linked article has the formula for the discriminant of a cubic.)

This gives us that $\Delta = \pm 9$. Thus, $$V_{p+2} = \dfrac{-3 \pm 9}{2} = -6 \text{ or } 3,$$ as you observed.

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For $p = 2$, one may do this "by hand": Since $\aa, \bb, \cc$ are all roots of $f$, they also satisfy $$X^{4} = 3X^{2} + X.$$ Thus, $V_{4} = V_{2} + V_{1}$. We have $V_{2} = V_{p} = V_{1} = 0$, so $V_{4} = 0$. (Note that $-6 \equiv 0 \bmod 2$.)