International Mathematics Competition

We show that $p=3$ is the only prime that satisfies the condition. Let $f(x)=x^{3}-3 x+1$. 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 $2 \cos 2 t=(2 \cos t)^{2}-2$, 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 $g(x)=\frac{f(x)}{x-a}=\frac{f(x)-f(a)}{x-a}=x^{2}+a x+\left(a^{2}-3\right)$. We claim that $b=a^{2}-2$ is a root of $g(x)$. Indeed, $g(b)=\left(a^{2}-2\right)^{2}+a\left(a^{2}-2\right)+\left(a^{2}-3\right)=(a+1) \cdot f(a)=0$. 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.