2019 USAMO Problems/Problem 1
Problem
Let
be the set of positive integers. A function
satisfies the equation
for all positive integers
. Given this information, determine all possible values of
.
Solution
Let
denote the result when
is applied to
, where
.
If
, then
and
since
.
Therefore,
is injective. It follows that
is also injective.
Lemma 1: If
and
, then
.
Proof:
which implies
by injectivity of
.
Lemma 2: If
, and
is odd, then
.
Proof:
Let
. Since
,
. So,
.
.
Since
,
This proves Lemma 2.
I claim that
for all odd
.
Otherwise, let
be the least counterexample.
Since
, either
, contradicted by Lemma 1 since
is odd and
.
, also contradicted by Lemma 1 by similar logic.
and
, which implies that
by Lemma 2.
This proves the claim.
By injectivity,
is not odd.
I will prove that
can be any even number,
. Let
, and
for all other
. If
is equal to neither
nor
, then
. This satisfies the given property.
If
is equal to
or
, then
since
is even and
. This satisfies the given property.
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See also
| 2019 USAMO (Problems • Resources) | ||
| First Problem | Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||