There is only one such function: $f(n)=n$. Of course \[f(k \cdot 1)=f(k) \cdot f(1)\] so $f(1)=1$. Now we prove by induction that $f(2^{k})=2^{k}$. Indeed $f(2)=2$ and \[f(2^{k+1})=f(2^{k}) \cdot f(2)= 2^{k+1}\] Now as the function is increasing, and we have \[2^{k+1}-2^{k}=f(2^{k+1})-f(2^{k})\] we get $f(n)=n$ for every $n \in N$

Other solution. We prove by induction that $ f(n)=n$ $ n=1,2$ then it is true. Suppose $ f(k)=k,\forall k=1,..,k$ We prove that $ f(k+1)=k+1$ Case 1 $ k+1$ is a prime then $ k+2$ is a composite . Suppose $ k+2=ab$ where $ a,b>1$ So $ f(k+2)=f(a)f(b)=ab=k+2$ But $ f(k)<f(k+1)<f(k+2)$ so $ f(k+1)=k+1$ Case 2 $ k+1=ab$ where $ a,b>1$ Then easy to check $ f(k+1)=k+1$ So $ f(n)=n,\forall n\in N$ Problem 2 $ f: N\to N$ $ f(mn)=f(m)f(n),\forall \gcd(m,n)=1$ $ f(2)=2,f(3)=3$ $ f(n+1)>(n)$ To solve this problem we use result: If $ f(n)=n,f(m)=m$ where $ m>n$ then $ f(k)=k,\forall k\in[n,m]$ So induction as above solution we get result :$ f(n)=n,\forall n\in N$

jastrzab wrote: There is only one such function: $ f(n) = n$. Of course \[ f(k \cdot 1) = f(k) \cdot f(1) \] so $ f(1) = 1$. Now we prove by induction that $ f(2^{k}) = 2^{k}$. Indeed $ f(2) = 2$ and \[ f(2^{k + 1}) = f(2^{k}) \cdot f(2) = 2^{k + 1} \] Now as the function is increasing, and we have \[ 2^{k + 1} - 2^{k} = f(2^{k + 1}) - f(2^{k}) \] we get $ f(n) = n$ for every $ n \in N$ I don't understand - how is the conclusion that f(n) = n reached. Could someone clarify it to me please?

The values $ f(2^k + a), a = 1, 2, ... 2^k - 1$ are strictly increasing and between $ 2^k$ and $ 2^{k+1}$, so they must take on each value between exactly once and in increasing order.

Peter wrote: Find all functions $ f: \mathbb{N} \to \mathbb{N}$ such that for all $ m,n\in \mathbb{N}$: $ f(2) = 2$, $ f(mn) = f(m)f(n)$, $ f(n + 1) > f(n)$. Ok,how about Find all functions $ f: \mathbb{N} \to \mathbb{N}$ such that for all $ m,n\in \mathbb{N}$: $ f(mn) = f(m)f(n)$, $ f(n + 1) > f(n)$.

Let try

Allnames' problem can be solved analogously by the hypothesis that $ f(n)=n^a$..

Allnames wrote: Find all functions $ f: \mathbb{N} \to \mathbb{N}$ such that for all $ m,n\in \mathbb{N}$: $ f(mn) = f(m)f(n)$, $ f(n + 1) > f(n)$.

Let try In fact, it suffices to have $ f(mn)=f(m)f(n)$ for coprime $ m,n$ only. It was some China TST, I think.

Proof without induction: We have one trivial solution $f(n)=n$. Now suppose that for some $n, f(n)$ is different form $n$. Let $k$ be the smallest such number, and let $f(k)=l$, for some $l>k$. Obviously $k$ has to be prime (if not, just take it's factorisation, and we get $f(k)=k$ because for all primes $p$ less than $k$ we have $f(p)=p$). So $k+1$ isn't prime ($k$ isn't $2$). Factorize $k+1$ and we easily get $f(k+1)=k+1$, and $k+1=f(k+1)>f(k)=l>k$. Contradiction.