Difference between revisions of "2019 USAMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
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− | Let <math>f^r(x)</math> denote the | + | 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>\ | + | <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) | + | 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> |
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− | Therefore, <math>f</math> is injective. | + | <math>\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2</math> |
− | <math> | + | |
− | Lemma 1: If <math>f^r(b)=a</math> and <math>f(a)=a</math>, then b=a | + | <math>\implies p=\pm q</math> |
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+ | <math>\implies p=q</math> since <math>p,q>0</math>. | ||
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+ | Therefore, <math>f</math> is injective. It follows that <math>f^r</math> is also injective. | ||
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+ | Lemma 1: If <math>f^r(b)=a</math> and <math>f(a)=a</math>, then <math>b=a</math>. | ||
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Proof: | Proof: | ||
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− | + | <math>f^r(b)=a=f^r(a)</math> which implies <math>b=a</math> by injectivity of <math>f^r</math>. | |
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Lemma 2: If <math>f^2(m)=f^{f(m)}(m)=m</math>, and <math>m</math> is odd, then <math>f(m)=m</math>. | Lemma 2: If <math>f^2(m)=f^{f(m)}(m)=m</math>, and <math>m</math> is odd, then <math>f(m)=m</math>. | ||
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Proof: | Proof: | ||
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− | Let <math>f(m)=k</math>. Since <math>f^2(m)=m</math>, <math>f(k)=m</math>. So, <math>f^2(k)=k</math>. <math>\newline f^2(k)\cdot f^{f(k)}(k)=k^2</math>. | + | Let <math>f(m)=k</math>. Since <math>f^2(m)=m</math>, <math>f(k)=m</math>. So, <math>f^2(k)=k</math>. <math>\newline f^2(k)\cdot f^{f(k)}(k)=k^2</math>. |
− | Since <math>k\neq0</math>, <math> | + | |
− | <math>\ | + | Since <math>k\neq0</math>, <math>f^{f(k)}(k)=k</math> |
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− | <math>\ | + | <math>\implies f^m(k)=k</math> |
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− | <math>\ | + | <math>\implies f^{gcd(m, 2)}(k)=k</math> |
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− | + | <math>\implies f(k)=k</math> | |
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This proves Lemma 2. | This proves Lemma 2. | ||
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I claim that <math>f(m)=m</math> for all odd <math>m</math>. | I claim that <math>f(m)=m</math> for all odd <math>m</math>. | ||
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Otherwise, let <math>m</math> be the least counterexample. | Otherwise, let <math>m</math> be the least counterexample. | ||
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− | 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) | + | <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) | + | <math>(2) f^{f(m)}(m)=k<m</math>, also contradicted by Lemma 1 by similar logic. |
− | <math> | + | |
− | (3) | + | <math>(3) f^2(m)=m</math> and <math>f^{f(m)}(m)=m</math>, which implies that <math>f(m)=m</math> by Lemma 2. |
This proves the claim. | This proves the claim. | ||
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By injectivity, <math>f(1000)</math> is not odd. | By injectivity, <math>f(1000)</math> is not odd. | ||
I will prove that <math>f(1000)</math> can be any even number, <math>x</math>. Let <math>f(1000)=x, f(x)=1000</math>, and <math>f(k)=k</math> for all other <math>k</math>. If <math>n</math> is equal to neither <math>1000</math> nor <math>x</math>, then <math>f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2</math>. This satisfies the given property. | I will prove that <math>f(1000)</math> can be any even number, <math>x</math>. Let <math>f(1000)=x, f(x)=1000</math>, and <math>f(k)=k</math> for all other <math>k</math>. If <math>n</math> is equal to neither <math>1000</math> nor <math>x</math>, then <math>f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2</math>. This satisfies the given property. | ||
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If <math>n</math> is equal to <math>1000</math> or <math>x</math>, then <math>f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2</math> since <math>f(n)</math> is even and <math>f^2(n)=n</math>. This satisfies the given property. | If <math>n</math> is equal to <math>1000</math> or <math>x</math>, then <math>f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2</math> since <math>f(n)</math> is even and <math>f^2(n)=n</math>. This satisfies the given property. | ||
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{{MAA Notice}} | {{MAA Notice}} |
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 |