IMC 2020 · Day 2 · P6

We show that $p = 3$ the only prime that satsfies the condition.

Let $f(x) = x^3 - 3x + 1$. As preparation, let's compute the roots of $f(x)$. By Cardano's formula, it can be seen that the roots are $$2 \operatorname{Re} \sqrt[3]{\frac{-1}{2} + \sqrt{\left( \frac{-1}{2} \right)^2 + \left( \frac{-3}{3} \right)^3}} = 2 \operatorname{Re} \sqrt[3]{\cos\frac{2\pi}{3} + i \sin\frac{2\pi}{3}} = \left\{ 2 \cos\frac{2\pi}{9}, 2 \cos\frac{4\pi}{9}, 2 \cos\frac{8\pi}{9} \right\}$$ where all three values of the complex cubic root were taken.

Notice that, by the trigonometric identity $2 \cos 2t = (2 \cos t)^2 - 2$, the map $\varphi(x) = x^2 - 2$ cyclically permutes the three roots. We will use this map to find another root of $f$, when it is considered over $\mathbb{F}_p$.

Suppose that $f(a) = 0$ for some $a \in \mathbb{F}_p$ and consider $$g(x) = \frac{f(x)}{x - a} = \frac{f(x) - f(a)}{x - a} = x^2 + a x + (a^2 - 3).$$ We claim that $b = a^2 - 2$ is a root of $g(x)$. Indeed, $$g(b) = (a^2 - 2)^2 + a (a^2 - 2) + (a^2 - 3) = (a + 1) \cdot f(a) = 0.$$ By Vieta's formulas, the other root of $g(x)$ is $c = -a - b = -a^2 - a + 2$.

If $f$ has a single root then the three roots must coincide, so $$a = a^2 - 2 = -a^2 - a + 2.$$ Here the quadratic equation $a = a^2 - 2$, or equivalently $(a + 1)(a - 2) = 0$, has two solutions, $a = -1$ and $a = 2$. By $f(-1) = f(2) = 3$, in both cases we have $0 = f(a) = 3$, so the only choice is $p = 3$.

Finally, for $p = 3$ we have $f(1) = -1$, $f(2) = 3$ and $f(3) = 19$, from these values only $f(2)$ is divisible by 3, so $p = 3$ satisfies the condition.

(outline) Define $f(x)$ and $g(x)$ like in Solution 1. The discriminant of $g(x)$ is $$\Delta_g = a^2 - 4 (a^2 - 3) = 12 - 3 a^2.$$ We show that $\Delta_g$ has a square root in $\mathbb{F}_p$.

Take two integers $k, m$ (to be determinated later) and consider $$\Delta_g = \Delta_g + (ka + m) f(a) = k a^4 + m a^3 - (3k + 1) a^2 + (k - 3m) a + (m + 12).$$ Our goal is to choose $k, m$ in such a way that the last expression is a complete square. Either by direct calculations or guessing, we can find that $k = m = 4$ works: $$\Delta_g = \Delta_g + (4a + 4) f(a) = 4 a^4 + 4 a^3 - 15 a^2 - 8a + 16 = (2 a^2 + a - 4)^2.$$ If $p \ne 2$ then we can conclude that $f(x)$ has either no or three roots, therefore $p$ is suitable if and only is $f(x)$ is a complete cube: $x^3 - 3x + 1 = (x - a)^3$. From Vieta's formulas $a^3 = 1$, so $a \ne 0$ and $3a = 0$, which is possible if $p = 3$.

For $p = 3$ we have $f(x) = (x + 1)^3$, so $p = 3$ is suitable.

The case $p = 2$ must be checked separately because the quadratic formula contains a division by 2. $f(1) = -1$ and $f(2) = 3$, so $p = 2$ is not suitable.

(outline) Assume $p > 3$; the cases $p = 2$ and $p = 3$ will be checked separately.

Let $f(x) = x^3 - 3x + 1$ and suppose that $a \in \mathbb{F}_p$ is a root of $f(x)$, and let $b, c \in \mathbb{F}_{p^2}$ be the other two roots. The discriminant $\Delta_f$ of $f(x)$ can be expressed by the elementary symmetric polynomials of $a, b, c$; it can be calculated that $$\Delta_f = (b - c)^2 (a - b)^2 (a - c)^2 = 81 = 9^2,$$ so $$(b - c)(a - b)(a - c) = \pm 9 \in \mathbb{F}_p.$$ Notice that $\Delta_f \ne 0$, so the three roots are distinct.

Either $b, c \in \mathbb{F}_p$ or $b, c$ are conjugate elements in $\mathbb{F}_{p^2} \setminus \mathbb{F}_p$, we have $(a - b)(a - c) \in \mathbb{F}_p$, so $b - c = \frac{(b-c)(a-b)(a-c)}{(a-b)(a-c)} \in \mathbb{F}_p$. From Vieta's formulas we have $b + c \in \mathbb{F}_p$ as well; since $p \ne 2$, it follows that $b, c \in \mathbb{F}_p$. Now $f(x)$ has three distinct roots in $\mathbb{F}_p$, so $p$ cannot be suitable.

$p = 2$ does not satisfies the condition because both $f(1) = -1$ and $f(2) = 3$ are odd. $p = 3$ is suitable, because $f(2) = 3$ is divisible by 3 while $f(1) = -1$ and $f(3) = 19$ are not.