If a prime natural number $p\neq 3$ divides $a^3-3a+1$ for some integer $a$, then $p\equiv \pm1\pmod{9}$.

Question

$\textbf{Problem:}$ Let $a$ be a positive integer and $p$ a prime divisor of $a^3-3a+1$, with $p \neq 3$. Prove that $p$ is of the form $9k+1$ or $9k-1$, where $k$ is an integer.

I tried to complete the cube which didn't turn out to be anything good. If the condition asked for $p$ to be only of the form $9k+1$, I would try to show that $a$ has order $9 \pmod{p}$. But the given condition seems somewhat odd to me. So, overall I could hardly make any real progress.

Any kind of hint or solutions are appreciated.

Answer 1

Without all details: If $\zeta$ is a primitive $9$th root of unity then $x^3-3x+1$ is the minimal polynomial of $\alpha = \zeta + \zeta^{-1}$. So if $x^3-3x+1$ has a root $\beta \pmod p$ for some prime then $\mathbb F_{p^2}$ has a root of $x^2-\beta x+1$ and if $p \neq 3$ then that root is a primitive $9$th root of unity. That implies that $9 \mid p^2-1$ since $\lvert \mathbb F_{p^2}^{\ast} \rvert = p^2-1$.

Answer 2

Let $f(x)=x^3-3x+1$. Then, $$g(t):=t^3\,f\left(t+\frac1t\right)=t^3\,\left(\left(t+\frac1t\right)^3-3\left(t+\frac1t\right)+1\right)\,.$$ Hence, $$g(t)=t^6+t^3+1=\Phi_9(t)\,,$$ where $\Phi_n$ is the $n$-th cyclotomic polynomial for each positive integer $n$.

Let $p\neq 3$ be a prime natural number. We work in $\mathbb{F}_p$. Suppose that $f(a)=0$ for some $a\in\mathbb{F}_p$. We consider the equation $$t+\frac{1}{t}=a\,.\tag{*}$$ Suppose that this equation has a solution $t=b$ for some $t\in\mathbb{F}_p\setminus\{0\}$, then $\Phi_9(b)=0$. As $t^9-1$ is divisible by $\Phi_9(t)$, we conclude that $$b^9-1=0\,.$$ However, $$b^{p-1}-1=0$$ as well. Consequently, if $d=\gcd(9,p-1)$, then $$b^{d}-1=0\,.$$ It is easily seen that $d\neq 1$ and $d\neq 3$; otherwise $b^3=1$, whence $$0=\Phi_9(b)=b^6+b^3+1=(b^3)^2+(b^3)+1=1^2+1+1=3\,,$$ contradicting the assumption that $p\neq 3$. Thus, $d=9$ implying that $9\mid p-1$.

Suppose now that (*) has a solution $t=b$, where $b\in\mathbb{F}_{p^2}\setminus\mathbb{F}_p$. Since $\Phi_9(b)=0$, we conclude as before that $$b^9-1=0\,.$$ However, since $b\in\mathbb{F}_{p^2}\setminus\{0\}$, we have $$b^{p^2-1}-1=0\,.$$ Therefore, $$b^{d}-1=0\,,$$ where $d:=\gcd(9,p^2-1)$. Using the same argument as the previous paragraph, $d=1$ and $d=3$ are ruled out. Therefore, $d=9$, making $9\mid p^2-1=(p-1)(p+1)$. Since $3$ divides exactly one of the numbers $p-1$ and $p+1$, we conclude that $9\mid p-1$ or $9\mid p+1$, establishing the claim.

Conversely, let $p$ be a prime natural number such that $p\equiv \pm1 \pmod{9}$. Then, $9\mid p^2-1$. Therefore, $\Phi_9(t)$ is a factor of $t^{p^2-1}-1$. Thus, $\Phi_9(t)$ splits into linear factors in $\mathbb{F}_{p^2}$. Let $b_1,b_2,b_3,b_1^{-1},b_2^{-1},b_3^{-1}$ be the six roots of $\Phi_9(t)$ in $\mathbb{F}_{p^2}\setminus\{0\}$. Then, the polynomial $f(x)=x^3-3x+1\in\mathbb{F}_p[x]$ has three roots $b_1+b_1^{-1}$, $b_2+b_2^{-1}$, and $b_3+b_3^{-1}$ in $\mathbb{F}_{p^2}$. Consequently, $f(x)$ cannot be irreducible over $\mathbb{F}_p$. Ergo, $f(x)$ has a root $a\in\mathbb{F}_p$. Therefore, we have the following proposition.

Proposition. Let $p\neq 3$ be a prime natural number. There exists an integer $a$ such that $a^3-3a+1$ is divisible by $p$ if and only if $p\equiv \pm1\pmod{9}$, or equivalently, $p\equiv \pm1\pmod{18}$.

Remark. In the case where $p\equiv \pm1\pmod{9}$, it can be easily seen that $f(x)$ has exactly three distinct roots in $\mathbb{F}_p$. This is because the discriminant of $f(x)$ is $81\not\equiv 0\pmod{p}$, and the roots of $f(x)$ are of the form $a$, $h(a)$, and $h\big(h(a)\big)$ for some $a\in\mathbb{F}_p$, where $h(x):=x^2-2$. See this related question: Expressing the roots of a cubic as polynomials in one root.