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Minimum possible value of $f(2007)$ where $f(m f(n)) = n f(m)$, $m,nin Bbb N$

Proof of minimum value of x+y when xy = afind $left( fracxx+y right)^2007 + left( fracyx+y right)^2007$Minimum sum of the squaresA function $f$ satisfies the condition $f[f(x) - e^x] = e + 1$ for all $x in Bbb R$.AM-GM Inequality concept challenged!$a in (0,1]$ satisfies $a^2008 -2a +1 = 0$ and we define $S$ as $S=1+a+a^2+a^3…a^2007$. The sum of all possible value(s) of $S$ is?Find possible value of $g(3)$Sum of $2008$ consecutive positive integersfind the maximum possible value of $n_9-n_1$Finding the minimum value.

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

9

$begingroup$

If $f$ is from positive integers to positive integers and satisfies

$f(m f(n)) = n f(m)$ then find the minimum possible value of $f(2007)$.

My work so far:

• 
  $f(1) = 1$ . Proof:

Suppose $f(1) = k neq 1$. Then consider $f(f(2)) = 2f(1) = 2k$.
$f(2) = f(2f(1)) = f(2k),$ $f(f(2)) = 2k,$ but the above also implies
that $f(f(2)) = f(f(2k)) = 2k^2$, a contradiction. Thus $f(1) = 1$

• I further found that if $f(a) = b$, then $f(a^x b^y) = a^y b^x$.

algebra-precalculus functional-equations integers multiplicative-function

share|cite|improve this question

edited Jul 4 at 9:13

smci

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asked Jul 3 at 18:44

doingmathdoingmath

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you factored $2007$?
  $endgroup$
  – Servaes
  Jul 3 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also note that $f(1)=1$ implies that $f(f(n))=n$ for all $n$, and so $f$ is bijective.
  $endgroup$
  – Servaes
  Jul 3 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At last... A problem where it matters whether $0 in mathbbN$. So, do you have $0 in mathbbN$ or not?
  $endgroup$
  – Eric Towers
  Jul 4 at 8:09
  

  
  
  
  

add a comment | 

9

$begingroup$

If $f$ is from positive integers to positive integers and satisfies

$f(m f(n)) = n f(m)$ then find the minimum possible value of $f(2007)$.

My work so far:

• 
  $f(1) = 1$ . Proof:

Suppose $f(1) = k neq 1$. Then consider $f(f(2)) = 2f(1) = 2k$.
$f(2) = f(2f(1)) = f(2k),$ $f(f(2)) = 2k,$ but the above also implies
that $f(f(2)) = f(f(2k)) = 2k^2$, a contradiction. Thus $f(1) = 1$

• I further found that if $f(a) = b$, then $f(a^x b^y) = a^y b^x$.

algebra-precalculus functional-equations integers multiplicative-function

share|cite|improve this question

edited Jul 4 at 9:13

smci

34822 silver badges1111 bronze badges

asked Jul 3 at 18:44

doingmathdoingmath

7433 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you factored $2007$?
  $endgroup$
  – Servaes
  Jul 3 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also note that $f(1)=1$ implies that $f(f(n))=n$ for all $n$, and so $f$ is bijective.
  $endgroup$
  – Servaes
  Jul 3 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At last... A problem where it matters whether $0 in mathbbN$. So, do you have $0 in mathbbN$ or not?
  $endgroup$
  – Eric Towers
  Jul 4 at 8:09
  

  
  
  
  

add a comment | 

$begingroup$

If $f$ is from positive integers to positive integers and satisfies

$f(m f(n)) = n f(m)$ then find the minimum possible value of $f(2007)$.

My work so far:

• 
  $f(1) = 1$ . Proof:

Suppose $f(1) = k neq 1$. Then consider $f(f(2)) = 2f(1) = 2k$.
$f(2) = f(2f(1)) = f(2k),$ $f(f(2)) = 2k,$ but the above also implies
that $f(f(2)) = f(f(2k)) = 2k^2$, a contradiction. Thus $f(1) = 1$

• I further found that if $f(a) = b$, then $f(a^x b^y) = a^y b^x$.

algebra-precalculus functional-equations integers multiplicative-function

share|cite|improve this question

edited Jul 4 at 9:13

smci

34822 silver badges1111 bronze badges

asked Jul 3 at 18:44

doingmathdoingmath

7433 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 4 at 9:13

smci

34822 silver badges1111 bronze badges

asked Jul 3 at 18:44

doingmathdoingmath

7433 bronze badges

share|cite|improve this question

edited Jul 4 at 9:13

smci

34822 silver badges1111 bronze badges

asked Jul 3 at 18:44

doingmathdoingmath

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edited Jul 4 at 9:13

smci

34822 silver badges1111 bronze badges

edited Jul 4 at 9:13

smci

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asked Jul 3 at 18:44

doingmathdoingmath

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asked Jul 3 at 18:44

doingmathdoingmath

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2 Answers
2

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oldest

votes

4

$begingroup$

You've already shown that $f(1)=1$. Plugging in $m=1$ it then follows that $f(f(n))=n$ for all $n$. In particular $f$ is bijective. Now for arbitrary $m$ and $n$, setting $k:=f(n)$ we see that $f(k)=f(f(n))$ and so
$$forall m,n: f(mf(n))=nf(m)qquadLeftrightarrowqquad forall k,m: f(km)=f(k)f(m),$$
which shows that $f$ is completely multiplicative. In particular this means that $f(2007)=f(3^2)f(223)$ because $2007=3^2times223$.

Now let $p$ be any prime number, and suppose $f(p)=ab$ for positive integers $a$ and $b$. Then
$$p=f(f(p))=f(ab)=f(a)f(b),$$
so without loss of generality $f(a)=1$. Then $a=f(f(a))=f(1)=1$ and so $f(p)$ is also prime. This means that $f$ permutes the set of primes. Because $f(f(p))=p$ for all primes this means $f$ is determined entirely by a permutation of order $2$ of the set of prime numbers.

I leave it to you to verify that conversely, every permutation of order $2$ of the set of prime numbers determines a completely multiplicative function that satisfies the functional equation. From there it is easy to verify that the minimum value of $f(2007)$ is $2times3^2=18$.

share|cite|improve this answer

edited Jul 3 at 19:23

answered Jul 3 at 19:21

ServaesServaes

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've not been able to follow the OP's line of reasoning when he writes that $f(f(2))=f(f(2k))=2k^2.$ How did he do that?
  $endgroup$
  – Adrian Keister
  Jul 3 at 19:23
  

  
  
  
  


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  @AdrianKeister Because $f(f(2k))=f(1cdot f(2k))=2kcdot f(1)=2kcdot k$.
  $endgroup$
  – Servaes
  Jul 3 at 19:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see, thanks!
  $endgroup$
  – Adrian Keister
  Jul 3 at 19:25
  

  
  
  
  


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  $begingroup$
  How did you conclude that $f(p)$ is also prime?
  $endgroup$
  – doingmath
  Jul 4 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @doingmath By starting off with positive integers $a$ and $b$ such that $f(p)=ab$. Then $$p=f(f(p))=f(ab)=f(a)f(b),$$ and so because $p$ is prime it follows that either $f(a)=1$ or $f(b)=1$. This implies that either $$a=f(f(a))=f(1)=1qquadtext or qquad b=f(f(b))=f(1)=1,$$ which means precisely that $f(p)$ is prime (or $f(p)=1$, but this is impossible because $f$ is bijective and $f(1)=1$).
  $endgroup$
  – Servaes
  Jul 4 at 18:31
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Let $a=f(1)neq 0$.

• Putting $m=1$ we get $f(f(n)) =an$ so $f$ is injective.


• For $n=1$ we get $f(ma) = f(m)$ so we have $ma=m$ so $a=1$ and thus $f(f(n)) =n$.

If we put $n=f(k)$ we get $f(mk)=f(k)f(m)$ so $f$ is multiplicative. Each such function is uniqely determined by the pictures of primes. We have $$f(2007) = f(3)^2f(223)$$

$f(3)=p$ and $f(223)=q$ are primes. Since $$f(p)=f(f(3))=3$$ and $$f(q)=f(f(223))=223$$ the minimum will be if we take $p=3$ and $q=2$.
so the answer is $18$?

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edited Jul 3 at 19:32

answered Jul 3 at 18:55

AquaAqua

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$endgroup$

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  $begingroup$
  Yes, $f$ is determined entirely by an involution on the set of primes.
  $endgroup$
  – Thomas Andrews
  Jul 3 at 19:27
  

  
  
  
  


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  $begingroup$
  But does there exist a function $f$ with $f(3)=3$ and $f(223)=2$ that satisfies the functional equation?
  $endgroup$
  – Servaes
  Jul 3 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/…
  $endgroup$
  – Aqua
  Jul 3 at 19:30
  
  

  
  
  
  


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  $begingroup$
  @Aqua What is your point? The fact that such a function is completely determined by its values at the prime numbers, does not mean that any choice of values at the prime numbers will yield a function satisfying the functional equation.
  $endgroup$
  – Servaes
  Jul 3 at 19:36
  
  

  
  
  
  


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  $begingroup$
  As Thomas Andrews notes in the comment above $f$ must induce an involution of the set of primes. But that still leaves the question; does every $f$ induced by an involution of the set of primes satisfy the functional equation?
  $endgroup$
  – Servaes
  Jul 3 at 19:38
  
  

  
  
  
  

 | 
show 2 more comments

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4

$begingroup$

You've already shown that $f(1)=1$. Plugging in $m=1$ it then follows that $f(f(n))=n$ for all $n$. In particular $f$ is bijective. Now for arbitrary $m$ and $n$, setting $k:=f(n)$ we see that $f(k)=f(f(n))$ and so
$$forall m,n: f(mf(n))=nf(m)qquadLeftrightarrowqquad forall k,m: f(km)=f(k)f(m),$$
which shows that $f$ is completely multiplicative. In particular this means that $f(2007)=f(3^2)f(223)$ because $2007=3^2times223$.

Now let $p$ be any prime number, and suppose $f(p)=ab$ for positive integers $a$ and $b$. Then
$$p=f(f(p))=f(ab)=f(a)f(b),$$
so without loss of generality $f(a)=1$. Then $a=f(f(a))=f(1)=1$ and so $f(p)$ is also prime. This means that $f$ permutes the set of primes. Because $f(f(p))=p$ for all primes this means $f$ is determined entirely by a permutation of order $2$ of the set of prime numbers.

I leave it to you to verify that conversely, every permutation of order $2$ of the set of prime numbers determines a completely multiplicative function that satisfies the functional equation. From there it is easy to verify that the minimum value of $f(2007)$ is $2times3^2=18$.

share|cite|improve this answer

edited Jul 3 at 19:23

answered Jul 3 at 19:21

ServaesServaes

34.8k44 gold badges4444 silver badges104104 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've not been able to follow the OP's line of reasoning when he writes that $f(f(2))=f(f(2k))=2k^2.$ How did he do that?
  $endgroup$
  – Adrian Keister
  Jul 3 at 19:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @AdrianKeister Because $f(f(2k))=f(1cdot f(2k))=2kcdot f(1)=2kcdot k$.
  $endgroup$
  – Servaes
  Jul 3 at 19:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see, thanks!
  $endgroup$
  – Adrian Keister
  Jul 3 at 19:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you conclude that $f(p)$ is also prime?
  $endgroup$
  – doingmath
  Jul 4 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @doingmath By starting off with positive integers $a$ and $b$ such that $f(p)=ab$. Then $$p=f(f(p))=f(ab)=f(a)f(b),$$ and so because $p$ is prime it follows that either $f(a)=1$ or $f(b)=1$. This implies that either $$a=f(f(a))=f(1)=1qquadtext or qquad b=f(f(b))=f(1)=1,$$ which means precisely that $f(p)$ is prime (or $f(p)=1$, but this is impossible because $f$ is bijective and $f(1)=1$).
  $endgroup$
  – Servaes
  Jul 4 at 18:31
  
  

  
  
  
  

add a comment | 

4

$begingroup$

You've already shown that $f(1)=1$. Plugging in $m=1$ it then follows that $f(f(n))=n$ for all $n$. In particular $f$ is bijective. Now for arbitrary $m$ and $n$, setting $k:=f(n)$ we see that $f(k)=f(f(n))$ and so
$$forall m,n: f(mf(n))=nf(m)qquadLeftrightarrowqquad forall k,m: f(km)=f(k)f(m),$$
which shows that $f$ is completely multiplicative. In particular this means that $f(2007)=f(3^2)f(223)$ because $2007=3^2times223$.

Now let $p$ be any prime number, and suppose $f(p)=ab$ for positive integers $a$ and $b$. Then
$$p=f(f(p))=f(ab)=f(a)f(b),$$
so without loss of generality $f(a)=1$. Then $a=f(f(a))=f(1)=1$ and so $f(p)$ is also prime. This means that $f$ permutes the set of primes. Because $f(f(p))=p$ for all primes this means $f$ is determined entirely by a permutation of order $2$ of the set of prime numbers.

I leave it to you to verify that conversely, every permutation of order $2$ of the set of prime numbers determines a completely multiplicative function that satisfies the functional equation. From there it is easy to verify that the minimum value of $f(2007)$ is $2times3^2=18$.

share|cite|improve this answer

edited Jul 3 at 19:23

answered Jul 3 at 19:21

ServaesServaes

34.8k44 gold badges4444 silver badges104104 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've not been able to follow the OP's line of reasoning when he writes that $f(f(2))=f(f(2k))=2k^2.$ How did he do that?
  $endgroup$
  – Adrian Keister
  Jul 3 at 19:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @AdrianKeister Because $f(f(2k))=f(1cdot f(2k))=2kcdot f(1)=2kcdot k$.
  $endgroup$
  – Servaes
  Jul 3 at 19:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see, thanks!
  $endgroup$
  – Adrian Keister
  Jul 3 at 19:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you conclude that $f(p)$ is also prime?
  $endgroup$
  – doingmath
  Jul 4 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @doingmath By starting off with positive integers $a$ and $b$ such that $f(p)=ab$. Then $$p=f(f(p))=f(ab)=f(a)f(b),$$ and so because $p$ is prime it follows that either $f(a)=1$ or $f(b)=1$. This implies that either $$a=f(f(a))=f(1)=1qquadtext or qquad b=f(f(b))=f(1)=1,$$ which means precisely that $f(p)$ is prime (or $f(p)=1$, but this is impossible because $f$ is bijective and $f(1)=1$).
  $endgroup$
  – Servaes
  Jul 4 at 18:31
  
  

  
  
  
  

add a comment | 

$begingroup$

You've already shown that $f(1)=1$. Plugging in $m=1$ it then follows that $f(f(n))=n$ for all $n$. In particular $f$ is bijective. Now for arbitrary $m$ and $n$, setting $k:=f(n)$ we see that $f(k)=f(f(n))$ and so
$$forall m,n: f(mf(n))=nf(m)qquadLeftrightarrowqquad forall k,m: f(km)=f(k)f(m),$$
which shows that $f$ is completely multiplicative. In particular this means that $f(2007)=f(3^2)f(223)$ because $2007=3^2times223$.

Now let $p$ be any prime number, and suppose $f(p)=ab$ for positive integers $a$ and $b$. Then
$$p=f(f(p))=f(ab)=f(a)f(b),$$
so without loss of generality $f(a)=1$. Then $a=f(f(a))=f(1)=1$ and so $f(p)$ is also prime. This means that $f$ permutes the set of primes. Because $f(f(p))=p$ for all primes this means $f$ is determined entirely by a permutation of order $2$ of the set of prime numbers.

I leave it to you to verify that conversely, every permutation of order $2$ of the set of prime numbers determines a completely multiplicative function that satisfies the functional equation. From there it is easy to verify that the minimum value of $f(2007)$ is $2times3^2=18$.

share|cite|improve this answer

edited Jul 3 at 19:23

answered Jul 3 at 19:21

ServaesServaes

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$endgroup$

share|cite|improve this answer

edited Jul 3 at 19:23

answered Jul 3 at 19:21

ServaesServaes

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answered Jul 3 at 19:21

ServaesServaes

34.8k44 gold badges4444 silver badges104104 bronze badges

answered Jul 3 at 19:21

ServaesServaes

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4

$begingroup$

Let $a=f(1)neq 0$.

• Putting $m=1$ we get $f(f(n)) =an$ so $f$ is injective.


• For $n=1$ we get $f(ma) = f(m)$ so we have $ma=m$ so $a=1$ and thus $f(f(n)) =n$.

If we put $n=f(k)$ we get $f(mk)=f(k)f(m)$ so $f$ is multiplicative. Each such function is uniqely determined by the pictures of primes. We have $$f(2007) = f(3)^2f(223)$$

$f(3)=p$ and $f(223)=q$ are primes. Since $$f(p)=f(f(3))=3$$ and $$f(q)=f(f(223))=223$$ the minimum will be if we take $p=3$ and $q=2$.
so the answer is $18$?

share|cite|improve this answer

edited Jul 3 at 19:32

answered Jul 3 at 18:55

AquaAqua

55.8k1414 gold badges6868 silver badges136136 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, $f$ is determined entirely by an involution on the set of primes.
  $endgroup$
  – Thomas Andrews
  Jul 3 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But does there exist a function $f$ with $f(3)=3$ and $f(223)=2$ that satisfies the functional equation?
  $endgroup$
  – Servaes
  Jul 3 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/…
  $endgroup$
  – Aqua
  Jul 3 at 19:30
  
  

  
  
  
  


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  @Aqua What is your point? The fact that such a function is completely determined by its values at the prime numbers, does not mean that any choice of values at the prime numbers will yield a function satisfying the functional equation.
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  – Servaes
  Jul 3 at 19:36
  
  

  
  
  
  


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  As Thomas Andrews notes in the comment above $f$ must induce an involution of the set of primes. But that still leaves the question; does every $f$ induced by an involution of the set of primes satisfy the functional equation?
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  – Servaes
  Jul 3 at 19:38
  
  

  
  
  
  

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show 2 more comments

4

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Let $a=f(1)neq 0$.

• Putting $m=1$ we get $f(f(n)) =an$ so $f$ is injective.


• For $n=1$ we get $f(ma) = f(m)$ so we have $ma=m$ so $a=1$ and thus $f(f(n)) =n$.

If we put $n=f(k)$ we get $f(mk)=f(k)f(m)$ so $f$ is multiplicative. Each such function is uniqely determined by the pictures of primes. We have $$f(2007) = f(3)^2f(223)$$

$f(3)=p$ and $f(223)=q$ are primes. Since $$f(p)=f(f(3))=3$$ and $$f(q)=f(f(223))=223$$ the minimum will be if we take $p=3$ and $q=2$.
so the answer is $18$?

share|cite|improve this answer

edited Jul 3 at 19:32

answered Jul 3 at 18:55

AquaAqua

55.8k1414 gold badges6868 silver badges136136 bronze badges

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  Yes, $f$ is determined entirely by an involution on the set of primes.
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  – Thomas Andrews
  Jul 3 at 19:27
  

  
  
  
  


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  But does there exist a function $f$ with $f(3)=3$ and $f(223)=2$ that satisfies the functional equation?
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  – Servaes
  Jul 3 at 19:28
  

  
  
  
  


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  $begingroup$
  en.wikipedia.org/wiki/…
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  – Aqua
  Jul 3 at 19:30
  
  

  
  
  
  


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  $begingroup$
  @Aqua What is your point? The fact that such a function is completely determined by its values at the prime numbers, does not mean that any choice of values at the prime numbers will yield a function satisfying the functional equation.
  $endgroup$
  – Servaes
  Jul 3 at 19:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As Thomas Andrews notes in the comment above $f$ must induce an involution of the set of primes. But that still leaves the question; does every $f$ induced by an involution of the set of primes satisfy the functional equation?
  $endgroup$
  – Servaes
  Jul 3 at 19:38
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Let $a=f(1)neq 0$.

• Putting $m=1$ we get $f(f(n)) =an$ so $f$ is injective.


• For $n=1$ we get $f(ma) = f(m)$ so we have $ma=m$ so $a=1$ and thus $f(f(n)) =n$.

If we put $n=f(k)$ we get $f(mk)=f(k)f(m)$ so $f$ is multiplicative. Each such function is uniqely determined by the pictures of primes. We have $$f(2007) = f(3)^2f(223)$$

$f(3)=p$ and $f(223)=q$ are primes. Since $$f(p)=f(f(3))=3$$ and $$f(q)=f(f(223))=223$$ the minimum will be if we take $p=3$ and $q=2$.
so the answer is $18$?

share|cite|improve this answer

edited Jul 3 at 19:32

answered Jul 3 at 18:55

AquaAqua

55.8k1414 gold badges6868 silver badges136136 bronze badges

$endgroup$

share|cite|improve this answer

edited Jul 3 at 19:32

answered Jul 3 at 18:55

AquaAqua

55.8k1414 gold badges6868 silver badges136136 bronze badges

answered Jul 3 at 18:55

AquaAqua

55.8k1414 gold badges6868 silver badges136136 bronze badges

answered Jul 3 at 18:55

AquaAqua

55.8k1414 gold badges6868 silver badges136136 bronze badges