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Find all functions $f:N \rightarrow N$ such that $f(2)=2$ and for any two $m,n ; f(mn)=f(m)f(n)$ and $f(n+1)>f(n)$.

Hint:
The set $S_n = \{n^q | q \in \mathbb Q\}$ is dense $\forall n \in \mathbb N, n > 1$.

Hints for two different solutions:

See that \((m,n)=(2,1)\) gives \(f(1)=1\) and \((m,n)=(2,2)\) gives \(f(4)=4\). Now \[2=f(2)<f(3)<f(4)=4\] so \(f(3)=3\). Suppose \(f(n)=n\) for \(n\le 2k\). Then \[f(2k+2)=f\left(2(k+1)\right)=f(2)f(k+1)=2k+2.\] And so \[2k=f(2k)<f(2k+1)<f(2k+2)=2k+2\] hence \(f(2k+1)=2k+1\). Therefore \(\boxed{f(n)=n}\) for all \(n\in\mathbb{N}\) by strong induction.

Nirjhor wrote:The given is Cauchy's Functional Equation. Hence \(f(x)=x^n\) for any fixed \(n\in\mathbb{N}\). Now \(f(2)=2^n=2\) gives \(n=1\) so \(f(n)=n\).

Wrong. For example the following set of functions defines a solution to $f(mn) = f(m)f(n), f: \mathbb N \mapsto \mathbb N$.
\[f(\prod p_i ^{e_i}) = \prod f(p_i)^{e_i}\]
\[f(p_i) = F(p_i)\]
Where the function $F: \mathbb P \mapsto \mathbb P$ is injective.

*Mahi* wrote:

Nirjhor wrote:The given is Cauchy's Functional Equation. Hence \(f(x)=x^n\) for any fixed \(n\in\mathbb{N}\). Now \(f(2)=2^n=2\) gives \(n=1\) so \(f(n)=n\).

Wrong. For example the following set of functions defines a solution to $f(mn) = f(m)f(n), f: \mathbb N \mapsto \mathbb N$.
\[f(\prod p_i ^{e_i}) = \prod f(p_i)^{e_i}\]
\[f(p_i) = F(p_i)\]
Where the function $F: \mathbb P \mapsto \mathbb P$ is injective.

Didn't understand what you wanted to say. The equality \[f\left(\prod p_i^{e_1}\right)=\prod f\left(p_i\right)^{e_i}\] is obvious from the given multiplicativity condition. If you're saying \(F:\mathbb{P}\mapsto\mathbb{P}\) is an arbitrary injective function, how does that satisfy the conditions given?

(I've edited my comment and added an inductive solution)

For example, $F(2)=3, F(3)=2$, and $F(p)=p$ for all other prime $p$ yields a solution.

I was considering the constraint that \(f\) is strictly increasing.

And I meant the part "$f:\mathbb N \mapsto \mathbb N, f(mn)=f(m)f(n) \Rightarrow$ Cauchy" is wrong, and provided a counter example. Even if you add the strictly increasing part, you'd have to prove it differently (like you did in the edit).

Got it. So is there any other solution to the given problem?