find strictly increasing function $f:\mathbb{N}\rightarrow \mathbb{R}$ such that $f(2)=2$ and $f(x\cdot y)=f(x)\cdot f(y)$ when $\gcd(x,y)=1$

Question

The problem is as in title... I understand that, according to for example answers to this question, trivial solution is $f(x)=x$. However, I don't see is weakening condition $f(x\cdot y)=f(x)\cdot f(y)$ to not have to hold for all $x,y\in \mathbb{N}$, but only when $\gcd(x,y)=1$, making it possible to find any alternative solution?

Thanks.

Answer 1

You can define $f(n)$ as the greatest power of $2$ that divides $n$.

Answer 2

You can define $f(x)$ to be whatever you like in all prime powers except 2 (where it is already defined), and that would still be OK. Then use multiplicity to reconstruct $f$ in all other integer points.

• Trivial solution: $f(p^n)=p^n$.
• Solution of Professor Vector: $\begin{cases}f(2^n)=2,\\f(p^n)=1, p\ne2\end{cases}$
• Solution of José Carlos Santos: $\begin{cases}f(2^n)=2^n,\\f(p^n)=1, p\ne2\end{cases}$

There are plenty of other possibilities. How about letting $f(p^n)=n+1$ (which is consistent with $f(2)=2$)? That would be the divisor function $\sigma_0$.