2002 IMO Shortlist Problems/A3


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.


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

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


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

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.)

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


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