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Problem

Let $f : \mathbb{N}\to\mathbb{N}$ be a strictly increasing function such that $f(2) = 2$ and $f(mn) = f(m) \cdot f(n)$ for every relatively prime pair of positive integers $m$ and $n$. Prove that $f(n) = n$ for every positive integer $n$.

My approach

I tried to prove this using induction. For the base case, we need to find $f(3)$. We have $$f(3) \cdot f(5)<2f(9)<4f(5)$$ $$\implies f(3)<4$$ And since $f(3)>f(2)=2$, we have $f(3)=3$.
Now, we assume that $f(k-1)=k-1$ and $f(k)=k$. Since $k$ and $k-1$ are relatively prime, we have $f(k(k-1))=f(k)f(k-1)=k(k-1)$. But this does not prove.

So, how do I complete the proof?

Oshawott
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    It is strictly increasing. If you know $f(2) = 2$ and $f(10) = 10$, can you say something about the function in the interval (2,10)? Try to think how to apply this to end your solution. – AnilCh Jun 22 '21 at 10:46
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    You can actually know that $f(x) = x$ in that interval. Think about it this way: You have $3,4,5,6,7,8,9$ in that interval. The image of $f$ in that interval, by the inequality you have written, must be $3,4,5,6,7,8,9$. And you know $f(3) < f(4) < f(5) < f(6) < f(7) < f(8) < f(9)$. So neccesarily $f(x)$ must be exactly $x$ in the interval. You can argue formally this by the pigeonhole principle or by contradiction, but the idea is that. – AnilCh Jun 22 '21 at 11:00
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    This can be generalized: if you have $f(a) = a, f(b) = b$ and $f$ is strictly increasing over the natural numbers, then $f(x) = x$ in $[a,b]$. This little exercise allow you to finish your exercise, thanks to your last equality. – AnilCh Jun 22 '21 at 11:03
  • @AnilCh Iterating $k \mapsto k(k - 1)$ will be unbounded and thus, prove it for all of $\Bbb N$, so that would finish the problem. Maybe you could post it as an answer. – Aryaman Maithani Jun 22 '21 at 11:32
  • This is Erdos Theorem: Let $f:\textbf{Z}^+\rightarrow \textbf{R}$ be a multiplicvative function. If $f$ is non-decraesing, then exists $a\in\textbf{R}$ such that $f(n)=n^a$, $\forall n \in\textbf{Z}^{+}$. – Nikos Bagis Jun 22 '21 at 12:56
  • For proof see: https://www.jstor.org/stable/2322314 – Nikos Bagis Jun 22 '21 at 12:59
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    Does this answer your question? Let $(a_n)$ be a strictly increasing sequence of positive integers such that: $a_2 = 2$, $a_{mn} = a_m a_n$ for $m, n$ relatively prime.. The other post is basically the same as this one except it uses a sequence of positive integers $a_n$ instead of an integer function. – John Omielan Jun 22 '21 at 13:55
  • @NikosBagis This has $\mathbb{N}$ as range. – AnilCh Jun 22 '21 at 14:33
  • @AnilCh. $\textbf{Z}^{+}=\textbf{N}$. See also my link. – Nikos Bagis Jun 22 '21 at 14:40
  • @NikosBagis I don't know if I'm misreading something here, but I said range, not domain. You've written $R$, which makes the problem much, much harder than what it actually is. – AnilCh Jun 22 '21 at 15:21

1 Answers1

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Show this little lemma previously: If $f:\mathbb{N}\rightarrow \mathbb{N}$ is a strictly increasing function over naturals and we have $a,b\in \mathbb{N}$ such that $f(a) = a, f(b) = b$ then $f(x) = x $ for all naturals in the interval $[a,b]$.

Proof of the lemma: Let $S$ be the set of naturals $S = \{a+1, ..., b-1\}$. Because $f$ is strictly increasing, we have $a < f(a+1) < f(a+2) < ... < f(b-1) < b$. In particular this means $f(S) = S$. Furthermore, $f(a+1)$ is the minimum of all values, because all those must be different it must mean that $f(a+1) = a+1$ because that's the minimum of $S$. If it wasn't $a+1$ you can reach a contradiction, because there must be a number in $S$ with image $a+1$. You can finish by applying this result to $f(a+1) = a+1, f(b) = b$.

After that you just apply your induction solution with $f(2) = 2$ and $f(k(k-1)) = k(k-1)$.

AnilCh
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