Difference between revisions of "2019 USAMO Problems/Problem 1"
m (→Solution) |
m (→Solution) |
||
(One intermediate revision by the same user not shown) | |||
Line 5: | Line 5: | ||
− | Let <math>f^r(x)</math> denote the result when <math>f</math> is applied to <math>x</math> <math> | + | Let <math>f^r(x)</math> denote the result when <math>f</math> is applied to <math>f^{r-1}(x)</math>, where <math>f^1(x)=f(x)</math>. |
<math>\hfill \break \hfill \break</math> | <math>\hfill \break \hfill \break</math> | ||
If <math>f(p)=f(q)</math>, then <math>f^2(p)=f^2(q)</math> and <math>f^{f(p)}(p)=f^{f(q)}(q)</math> | If <math>f(p)=f(q)</math>, then <math>f^2(p)=f^2(q)</math> and <math>f^{f(p)}(p)=f^{f(q)}(q)</math> | ||
Line 35: | Line 35: | ||
Since <math>k\neq0</math>, <math>f^{f(k)}(k)=k</math> | Since <math>k\neq0</math>, <math>f^{f(k)}(k)=k</math> | ||
− | <math>f^m(k)=k</math> | + | <math>\implies f^m(k)=k</math> |
− | <math>f^{gcd(m, 2)}(k)=k</math> | + | <math>\implies f^{gcd(m, 2)}(k)=k</math> |
− | <math>f(k)=k</math> | + | <math>\implies f(k)=k</math> |
This proves Lemma 2. | This proves Lemma 2. | ||
Line 50: | Line 50: | ||
Since <math>f^2(m)\cdot f^{f(m)}(m)=m^2</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^2(k)=k</math>. |
<math>(2) f^{f(m)}(m)=k<m</math>, also contradicted by Lemma 1 by similar logic. | <math>(2) f^{f(m)}(m)=k<m</math>, also contradicted by Lemma 1 by similar logic. |
Latest revision as of 18:04, 5 April 2021
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.
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 |