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| ==Solution== | | ==Solution== |
− | \usepackage{amsmath}
| |
| | | |
| <math>\newline</math> | | <math>\newline</math> |
Line 52: |
Line 51: |
| <math>\newline</math> | | <math>\newline</math> |
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− | <math>\newline</math>
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− | Let <math>f^r(x)</math> denote the resulr when <math>f</math> is applied to <math>x</math> <math>r</math> times.
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− | <math>\newline\newline</math>
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− | 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)\newline\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2\newline\implies p=\pm q\newline\implies p=q</math>since <math>p,q>0</math>.
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− | \newline
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− | Therefore, <math>f</math> is injective.
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− | <math>\newline\newline</math>
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− | Lemma 1: If <math>f^r(b)=a</math> and <math>f(a)=a</math>, then b=a.
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− | <math>\newline\newline</math>
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− | Proof:
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− | <math>\newline</math>
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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.
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− | <math>\newline\newline</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>.
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− | <math>\newline\newline</math>
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− | Proof:
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− | <math>\newline</math>
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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>.
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− | <math>\newline</math>
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− | Since <math>k\neq0</math>, <math>\newlinef^{f(k)}(k)=k</math>
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− | <math>\newline</math>
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− | <math>f^m(k)=k</math>
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− | <math>\newline</math>
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− | <math>f^{gcd(m, 2)}=k</math>
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− | <math>\newline</math>
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− | <math>m=f(k)=k</math>
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− | <math>\newline</math>
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− | This proves Lemma 2.
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− | <math>\newline\newline</math>
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− | I claim that <math>f(m)=m</math> for all odd <math>m</math>.
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− | <math>\newline</math>
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− | Otherwise, let <math>m</math> be the least counterexample.
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− | <math>\newline</math>
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− | Since <math>f^2(m)\cdot f^{f(m)}(m)</math>, either
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− | <math>\newline</math>
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− | (1) <math>f^2(m)=k<m</math>, contradicted by Lemma 1 since <math>f(k)=k</math>.
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− | <math>\newline</math>
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− | (2) <math>f^{f(m)}(m)=k<m</math>, also contradicted by Lemma 1.
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− | <math>\newline</math>
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− | (3) <math>f^2(m)=m</math> and <math>f^{f(m)}(m)=m</math>, which implies that <math>f(m)=m</math> by Lemma 2.
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− | This proves the claim.
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− | <math>\newline\newline</math>
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− | 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.
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− | <math>\newline</math>
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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.
| |
− | <math>\newline</math>
| |
| | | |
| | | |
Revision as of 20:46, 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 resulr when
is applied to
times.
If
, then
and
since
.
\newline
Therefore,
is injective.
Lemma 1: If
and
, then b=a.
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
, $\newlinef^{f(k)}(k)=k$ (Error compiling LaTeX. Unknown error_msg)
This proves Lemma 2.
I claim that
for all odd
.
Otherwise, let
be the least counterexample.
Since
, either
(1)
, contradicted by Lemma 1 since
.
(2)
, also contradicted by Lemma 1.
(3)
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