2002 IMO Shortlist Problems/A3
We note that the condition is equivalent to
Since , we may remove the factor to obtain
We can denote and as and , respectively, rewriting this as
We claim that can only assume finitely many values. We note that , so , which brings us to
which is clearly true for at most finitely many integer values of .
We denote by . The condition is then equivalent to , or . But can only assume finitely many values, so for some value of , are quartic polynomials equivalent at infinitely many points and are therefore equivalent.
Now, if , then we have , so is an even function. Since is divisible by , it must therefore also be divisible by , implying that is divisible by , which means that 0 is a root of , as desired.
If , then we have , so , implying that is equal to 0 since . Thus is the desired root of .
Consider the function . If has roots , then is bounded on the interval , and injective on each of the intervals . Because each value of yields at most finitely many solutions , there are at most finitely many solutions such that and have the same sign.
Since , if , then each value of yields at most one solution. But for arbitrarily large values of , and have the same sign, making impossible if and have different signs. Thus there must be some integer such that . (Indeed, if , then any pair of integers with sum is a solution.)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.