Difference between revisions of "2002 IMO Shortlist Problems/A3"

Revision as of 16:37, 1 January 2007

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.)