Difference between revisions of "2019 USAMO Problems/Problem 1"
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Otherwise, let <math>m</math> be the least counterexample. | Otherwise, let <math>m</math> be the least counterexample. | ||
− | Since <math>f^2(m)\cdot f^{f(m)}(m)</math>, either | + | Since <math>f^2(m)\cdot f^{f(m)}(m)=m^2</math>, either |
<math>(1) f^2(m)=k<m</math>, contradicted by Lemma 1 since <math>k</math> is odd and <math>f(k)=k</math>. | <math>(1) f^2(m)=k<m</math>, contradicted by Lemma 1 since <math>k</math> is odd and <math>f(k)=k</math>. |
Revision as of 00:24, 25 April 2019
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
times.
If
, then
and
since
.
Therefore, is injective.
Lemma 1: If and
, then
.
Proof:
which implies
by applying injectivity of
times.
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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |