2010 IMO Problems/Problem 3
Problem
Find all functions such that
is a perfect square for all
Author: Gabriel Carroll, USA
Solution
Suppose such function exist then:
Lemma 1)
Assume for contradiction that![]()
has to be a perfect square
but .
A square cannot be between 2 consecutive squares. Contradiction. Thus,![]()
Lemma 2) (we have show that it can't be 0)
Assume for contradiction, that.
Then there must exist a prime number such that
and
are in the same residue class modulo
.
If where
is not divisible by
.
If.
Consider ansuch that
![]()
, where
is not divisible by
![]()
If.
Consider ansuch that
![]()
, where
is not divisible by
![]()
At least one of,
is not divisible by
. Hence,
At least one of,
is divisible by an odd amount of
.
Hence, that number is not a perfect square.
If , then
,
.
, which is not perfect square because
is never a perfect square.
If , then
,
.
Thus,,
![]()