Difference between revisions of "2014 USAMO Problems/Problem 2"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | x^6f(x) = x^4(-x)^2f(-(-x))\\ | + | x^6f(x) &= x^4(-x)^2f(-(-x))\\ |
&= x^4f(-x)^2\\ | &= x^4f(-x)^2\\ | ||
&= f(x)^4 | &= f(x)^4 |
Revision as of 20:58, 24 March 2015
Problem
Let be the set of integers. Find all functions
such that
for all
with
.
Solution
Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.
Lemma 1: .
Proof: Assume the opposite for a contradiction. Plug in
(because we assumed that
),
. What you get eventually reduces to:
which is a contradiction since the LHS is divisible by 2 but not 4.
Then plug in into the original equation and simplify by Lemma 1. We get:
Then:
Therefore, must be 0 or
.
Now either is
for all
or there exists
such that
. The first case gives a valid solution. In the second case, we let
in the original equation and simplify to get:
But we know that
, so:
Since
is not 0,
is 0 for all
(including 0). Now either
is 0 for all
, or there exists some
such that
. Then
must be odd. We can let
in the original equation, and since
is 0 for all
, stuff cancels and we get:
[b]for
.[/b]
Now, let
and we get:
Now, either both sides are 0 or both are equal to
. If both are
then:
which simplifies to:
Since
and
is odd, both cases are impossible, so we must have:
Then we can let
be anything except 0, and get
is 0 for all
except
. Also since
, we have
, so
is 0 for all
except
. So
is 0 for all
except
. Since
,
. Squaring,
and dividing by
,
. Since
,
, which is a contradiction, so our only solutions are
and
.