International Mathematics Competition
for University Students
2020

IMC2020: Day 2, Problem 6

Problem 6. Find all prime numbers \(\displaystyle p\) for which there exists a unique \(\displaystyle a\in \{1,2,\ldots,p\}\) such that \(\displaystyle a^3-3a+1\) is divisible by \(\displaystyle p\).

Géza Kós, Loránd Eötvös University, Budapest

Solution 1. We show that \(\displaystyle p=3\) the only prime that satsfies the condition.

Let \(\displaystyle f(x)=x^3-3x+1\). As preparation, let's compute the roots of \(\displaystyle f(x)\). By Cardano's formula, it can be seen that the roots are

\(\displaystyle 2\mathrm{Re}\sqrt[3]{\frac{-1}{2}+\sqrt{ \bigg(\frac{-1}{2}\bigg)^2-\bigg(\frac{-3}{3}\bigg)^3}} = 2\mathrm{Re}\sqrt[3]{\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}} = \bigg\{ 2\cos\frac{2\pi}{9}, 2\cos\frac{4\pi}{9}, 2\cos\frac{8\pi}{9} \bigg\} \)

where all three values of the complex cubic root were taken.

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

Suppose that \(\displaystyle f(a)=0\) for some \(\displaystyle a\in\mathbb{F}_p\) and consider

\(\displaystyle g(x) = \frac{f(x)}{x-a} = \frac{f(x)-f(a)}{x-a} = x^2+ax+(a^2-3). \)

We claim that \(\displaystyle b=a^2-2\) is a root of \(\displaystyle g(x)\). Indeed,

\(\displaystyle g(b) = (a^2-2)^2+a(a^2-2)+(a^2-3) = (a+1)\cdot f(a) = 0. \)

By Vieta's formulas, the other root of \(\displaystyle g(x)\) is \(\displaystyle c=-a-b=-a^2-a+2\).

If \(\displaystyle f\) has a single root then the three roots must coincide, so

\(\displaystyle a = a^2-2 = -a^2-a+2. \)

Here the quadratic equation \(\displaystyle a=a^2-2\), or equivalently \(\displaystyle (a+1)(a-2)=0\), has two solutions, \(\displaystyle a=-1\) and \(\displaystyle a=2\). By \(\displaystyle f(-1)=f(2)=3\), in both cases we have \(\displaystyle 0=f(a)=3\), so the only choice is \(\displaystyle p=3\).

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

Solution 2 (outline) Define \(\displaystyle f(x)\) and \(\displaystyle g(x)\) like in Solution 1. The discriminant of \(\displaystyle g(x)\) is

\(\displaystyle \Delta_g = a^2-4(a^2-3) = 12-3a^2. \)

We show that \(\displaystyle \Delta_g\) has a square root in \(\displaystyle \mathbb{F}_p\).

Take two integers \(\displaystyle k,m\) (to be determinated later) and consider

\(\displaystyle \Delta_g = \Delta_g +(ka+m)f(a) = ka^4+ma^3-(3k+1)a^2+(k-3m)a+(m+12). \)

Our goal is to choose \(\displaystyle k,m\) in such a way that the last expression is a complete square. Either by direct calculations or guessing, we can find that \(\displaystyle k=m=4\) works:

\(\displaystyle \Delta_g = \Delta_g +(4a+4)f(a) = 4a^4+4a^3-15a^2-8a+16 = (2a^2+a-4)^2. \)

If \(\displaystyle p\ne2\) then we can conclude that \(\displaystyle f(x)\) has either no or three roots, therefore \(\displaystyle p\) is suitable if and only is \(\displaystyle f(x)\) is a complete cube: \(\displaystyle x^3-3x+1 = (x-a)^3\). From Vieta's formulas \(\displaystyle a^3=1\), so \(\displaystyle a\ne0\) and \(\displaystyle 3a=0\), which is possible if \(\displaystyle p=3\).

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

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

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

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

\(\displaystyle \Delta_f=(b-c)^2(a-b)^2(a-c)^2 = 81 = 9^2, \)

so

\(\displaystyle (b-c)(a-b)(a-c) = \pm9 \in \mathbb{F}_p. \)

Notice that \(\displaystyle \Delta_f\ne0\), so the three roots are distinct.

Either \(\displaystyle b,c\in \mathbb{F}_p\) or \(\displaystyle b,c\) are conjugate elements in \(\displaystyle \mathbb{F}_{p^2}\setminus\mathbb{F}_p\), we have \(\displaystyle (a-b)(a-c)\in\mathbb{F}_p\), so \(\displaystyle b-c=\frac{(b-c)(a-b)(a-c)}{(a-b)(a-c)}\in\mathbb{F}_p\). From Vieta's formulas we have \(\displaystyle b+c\in\mathbb{F}_p\) as well; since \(\displaystyle p\ne2\), it follows that \(\displaystyle b,c\in\mathbb{F}_p\). Now \(\displaystyle f(x)\) has three distinct roots in \(\displaystyle \mathbb{F}_p\), so \(\displaystyle p\) cannot be suitable.

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