(Arithmetic of functions) 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 $f(3)=3+k, k \geqslant 0$. By successive enclosures on small values, we show that $k=0$: indeed, $f(6)=2 k+6$ so $f(5) \leqslant 2 k+5$ and $f(10) \leqslant 4 k+10, f(9) \leqslant 4 k+9, f(18) \leqslant 8 k+18$. And $f(5) \geqslant 5+k$ so $k^{2}+8 k+15=(3+k)(5+k) \leqslant f(15) \leqslant f(18)-3 \leqslant 8 k+15 ;$ from this we deduce $k=0$.

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

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

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