2002 IMO Shortlist Problems/A3
Contents
[hide]Problem
Let be a cubic polynomial given by
, where
are integers and
. Suppose that
for infinitely many pairs
,
of integers with
. Prove that the equation
has an integer root.
Solutions
We note that the condition is equivalent to
.
Since , we may remove the factor
to obtain
,
or
.
We can denote and
as
and
, respectively, rewriting this as
.
Solution 1
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
.
Solution 2
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.