Find all strictly increasing functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(2)=2$ and for all $m, n$ that are coprime, $f(m n)=f(m) f(n)$.
By strict increase of $f, f(0)=0$ and $f(1)=1$. To multiply numbers that are coprime without using 1, we would like to know $f(3)$ for example. Let $f(3)=3+k, k \geq 0$. By successive enclosures on small values, we show that $k=0$: indeed, $f(6)=2k+6$ so $f(5) \leq 2k+5$ and $f(10) \leq 4k+10, f(9) \leq 4k+9, f(18) \leq 8k+18$. And, $f(5) \geq 5+k$ so $f(3+k)(5+k) \leq f(15) \leq f(18)-3$ and from this we deduce $k=0$.

Now, let's prove by induction on $n$ that $f(k)=k$ if $k \leq 2^{n}+1$. We have verified the initialization at $n=1$, and for the induction step:

$$
f\left(2^{n+1}+2\right)=f(2) f\left(2^{n}+1\right)=2 \cdot (n+1)+2
$$

so the $f(k), 2^{n}+2 \leq k \leq 2^{n+2}+2$ are between $2^{n}+2$ and $2^{n+2}+2$ in the broad sense, and the strict increase of the function ensures that $f(k)=k$ for all $k \leq 2^{n+1}+1$. This closes the induction, and allows us to assert that $f$ is the identity. Conversely, this function is obviously a solution to the problem.