Find all functions satisfying f(mn)=f(m)f(n), and m+n|f(m)+f(n)

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In summary, we are looking for all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ that satisfy the conditions $f(mn)=f(m)f(n)$ and $m+n | f(m)+f(n)$ for all $m,n \in \mathbb{N}$. Additionally, odd powers of $n$ also have this property, and it is unknown whether these are the only solutions.
  • #1
lfdahl
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Find all functions, $f: \mathbb{N}\rightarrow \mathbb{N}$, satisfying

\[f(mn)=f(m)f(n),\: \: \: and \: \: \: m+n \: \: |\: \: f(m)+f(n)\]for all $m,n \in \mathbb{N}$.
 
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  • #2
lfdahl said:
Find all functions, $f: \mathbb{N}\rightarrow \mathbb{N}$, satisfying

\[f(mn)=f(m)f(n),\: \: \: and \: \: \: m+n \: \: |\: \: f(m)+f(n)\]for all $m,n \in \mathbb{N}$.

by putting m = 1 we get $f(n) = f(1) f(n)$ so $f(1) = 1$

now 1 + n is a factor of 1 + f(n) so by inspection f(n) = n. I do not have a proof of the same
 
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  • #3
kaliprasad said:
by putting m = 1 we get $f(n) = f(1) f(n)$ so $f(1) = 1$

now 1 + n is a factor of 1 + f(n) so by inspection f(n) = n. I do not have a proof of the same
[sp]Odd powers of $n$ also have this property: $(mn)^k = m^kn^k$, and $(m+n)^k = (m+n)(m^{k-1} - m^{k-2}n + \ldots + n^k)$ if $k$ is an odd integer.

I have not thought about whether these are the only solutions.
[/sp]
 
  • #4
Opalg said:
[sp]Odd powers of $n$ also have this property: $(mn)^k = m^kn^k$, and $(m+n)^k = (m+n)(m^{k-1} - m^{k-2}n + \ldots + n^k)$ if $k$ is an odd integer.

I have not thought about whether these are the only solutions.
[/sp]

Right
 
  • #5
Opalg said:
[sp]Odd powers of $n$ also have this property: $(mn)^k = m^kn^k$, and $(m+n)^k = (m+n)(m^{k-1} - m^{k-2}n + \ldots + n^k)$ if $k$ is an odd integer.

I have not thought about whether these are the only solutions.
[/sp]

there is a typo in above

$(m+n)^k = (m+n)(m^{k-1} - m^{k-2}n + \ldots + n^k)$

should be

$m^k+n^k = (m+n)(m^{k-1} - m^{k-2}n + \ldots + n^{k-1})$
 
  • #6
Suggested solution:
As kaliprasad noted: $f(1) = 1$. So, $2n+1 \;\; | \;\; f(2n) + f(1) = f(2) \cdot f(n) + 1 \Rightarrow gcd(2n+1,f(2)) = 1$ for all $n$. This means, that $f(2)$ has no odd prime divisor, so $f(2) = 2^k$. Furthermore, $3 \;\; | 1 + f(2) = 1 + 2^k \Rightarrow k$ is odd (as Opalg noted). Also note, $f(2^m) = f(2) \cdot f(2) \cdot … \cdot f(2) = 2^{km}$ for all $m$. Now, for all $m$ and $n$, $2^m+n \;\; | \;\; f(2^{m}) + f(n) = 2^{km} + f(n)$. But $k$ is odd, so $2^m+n \;\; | \;\; 2^{km} + n^k$. Thus $2^m+n \;\; | \;\; f(n)-n^k$. But this is valid for all $m$, so $f(n) – n^k = 0 \Rightarrow f(n) = n^k.$
We have now proven, that $f(n) = n^k$ for all $n$ for some fixed odd number $k$. On the other hand, it is easily seen that functions of this form satisfy the given conditions.
 

FAQ: Find all functions satisfying f(mn)=f(m)f(n), and m+n|f(m)+f(n)

1. What is the purpose of finding all functions satisfying f(mn)=f(m)f(n)?

The purpose is to identify all possible functions that satisfy the given equation, which can help in solving related problems and understanding the properties of the functions involved.

2. How do you approach finding these functions?

The approach involves analyzing the given equation and using mathematical techniques such as substitution, induction, and contradiction to derive possible solutions.

3. Are there any specific restrictions or conditions for the functions?

Yes, the functions must satisfy the equation f(mn)=f(m)f(n) and m+n|f(m)+f(n). Additionally, the functions should be defined for all positive integer inputs.

4. Can there be multiple solutions for these functions?

Yes, there can be multiple solutions for the given equation. In fact, there are infinitely many functions that can satisfy the given conditions.

5. How can these functions be applied in real-world situations?

These functions can be applied in various mathematical and scientific fields, such as number theory, abstract algebra, and cryptography. They can also be used to model real-world phenomena and analyze their properties.

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