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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.

Crni
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  • There are lots of solutions. $f(2n)=2$ and $f(2n-1)=1$ for $n\ge1$ would be just one example. –  Oct 05 '17 at 09:21
  • I'm very sorry, forgot to mention an additional requirement and that is that the function has to be strictly increasing: $x>y\implies f(x)>f(y)$. I've edited the title accordingly now. Any suggestions for this case? – Crni Oct 05 '17 at 13:29
  • With that correction, I feel there will be only the "trivial" solution $f(n)=n$, but showing that will be very far from trivial. –  Oct 05 '17 at 14:05

2 Answers2

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

José Carlos Santos
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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 $\sigma_0$.

Ivan Neretin
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