2002 IMO Shortlist Problems/A3

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Problem

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

Solutions

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

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

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

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

or

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


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

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

Solution 1

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

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

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

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

Solution 2

Consider the function $\displaystyle xP(x)$. If $\displaystyle P(x)$ has roots $\lambda_1 \le \lambda_2 \le \lambda_3$, then $\displaystyle P(x)$ 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 $\displaystyle xP(x)$ yields at most finitely many solutions $\displaystyle x,y$, there are at most finitely many solutions $\displaystyle x,y$ such that $\displaystyle x$ and $\displaystyle y$ have the same sign.

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