Difference between revisions of "2019 USAMO Problems/Problem 1"
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− | Otherwise, set <math>a</math>, <math>b</math>, and <math>r</math> to a counterexample of the lemma, such that <math>r</math> is minimized. By injectivity, <math>f(a)=a\implies f(b)\neq a</math>, so <math>r\neq1</math>. If <math>f^n(b)=a</math>, then <math>f^n-1(f(b))=a</math> and <math>f(b)\neq a</math>, a counterexample that contradicts our assumption that <math>r</math> is minimized, proving Lemma 1. | + | Otherwise, set <math>a</math>, <math>b</math>, and <math>r</math> to a counterexample of the lemma, such that <math>r</math> is minimized. By injectivity, <math>f(a)=a\implies f(b)\neq a</math>, so <math>r\neq1</math>. If <math>f^n(b)=a</math>, then <math>f^{n-1}(f(b))=a</math> and <math>f(b)\neq a</math>, a counterexample that contradicts our assumption that <math>r</math> is minimized, proving Lemma 1. |
Revision as of 23:15, 24 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:
Otherwise, set ,
, and
to a counterexample of the lemma, such that
is minimized. By injectivity,
, so
. If
, then
and
, a counterexample that contradicts our assumption that
is minimized, proving Lemma 1.
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
.
, also contradicted by Lemma 1.
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 |