2002 IMO Shortlist Problems/A3

Problem

Let $P$ be a cubic polynomial given by $P(x) = ax^3 + bx^2 + cx + d$ , where $a, b, c, d$ are integers and $a \neq 0$ . Suppose that $xP(x) = yP(y)$ for infinitely many pairs $x$ , $y$ of integers with $x \neq y$ . Prove that the equation $P(x) = 0$ has an integer root.

Solutions

We note that the condition $xP(x) = yP(y)$ is equivalent to

$a(x^4 - y^4) + b(x^3 - y^3) + c(x^2 - y^2) + d(x - y) = 0$ .

Since $x \neq y$ , we may remove the factor $(x-y)$ to obtain

$a(x^3 + x^{2}y + xy^2 + y^3) + b(x^2 + xy + y^2) + c(x + y) + d = 0$ ,

or

$P(x+y) = xy\left[ 2a(x+y) + b \right]$ .

We can denote $x + y$ and $xy$ as $s$ and $t$ , respectively, rewriting this as

$P(s) = (2as + b)t$ .

Solution 1

We claim that $s$ can only assume finitely many values. We note that ${} s^2 - 4t = (x-y)^2 \ge 0$ , so $|t| < s^2 /4$ , which brings us to

$| as^3 + bs^2 + cs + d | \le \left| \frac{a}{2}s^3 + \frac{b}{4}s^2 \right|$ ,

which is clearly true for at most finitely many integer values of $s$ .

We denote $xP(x)$ by $Q(x)$ . The condition $xP(x) = yP(y)$ is then equivalent to $Q(x) = Q(y)$ , or $Q(s) = Q(r-s)$ . But $s$ can only assume finitely many values, so for some value of $s$ , $Q(x) = Q(s-x)$ are quartic polynomials equivalent at infinitely many points and are therefore equivalent.

Now, if $s = 0$ , then we have $Q(x) = Q(-x)$ , so $Q$ is an even function. Since $Q$ is divisible by $x$ , it must therefore also be divisible by $x^2$ , implying that $P$ is divisible by $x$ , which means that 0 is a root of $P$ , as desired.

If $s \neq 0$ , then we have $xP(x) = (s-x)P(s-x)$ , so $sP(s) = 0$ , implying that $P(s)$ is equal to 0 since $s \neq 0$ . Thus $s$ is the desired root of $P(s)$ .

Solution 2

Consider the function $xP(x)$ . If $P(x)$ has roots $\lambda_1 \le \lambda_2 \le \lambda_3$ , then $P(x)$ is bounded on the interval $\left[ \min (\lambda_1 , 0) , \max (\lambda_3 , 0) \right]$ , and injective on each of the intervals $(-\infty , \min (\lambda_1 , 0) ) , ( \max (\lambda_3 , 0 ) , \infty)$ . Because each value of $xP(x)$ yields at most finitely many solutions $x,y$ , there are at most finitely many solutions $x,y$ such that $x$ and $y$ have the same sign.

Since $P(s) = xy( 2as + b )$ , if $P(s) \neq 0$ , then each value of $s$ yields at most one solution. But for arbitrarily large values of $s$ , $P(s)$ and $(2as + b)$ have the same sign, making $P(s) = xy (2as + b)$ impossible if $x$ and $y$ have different signs. Thus there must be some integer $s$ such that $P(s) = 0$ . (Indeed, if $P(s) = 2as + b = 0$ , then any pair of integers with sum $s$ is a solution.)