ok ill post my solution: Now clearly $f(x)=x^2$ is a function that satisfies all the conditions, yielding $f(3)=9$ But with this conjecture we need to prove that $f(3)\neq7, 8$ Separate this into two cases. CASE 1 $f(3)=7$ $\implies 2\leq f(2) \leq 6$ We need to check that for each of these possible values of $f(2)$, we arrive at a contradiction. If $f(2) = 2$ $\implies f(4)=4, f(3)=7$. contradiction because $f(a)<f(b)$ if $a<b$ Now, if $f(2)=3$ $\implies f(9)=49, f(12)=63$ $\implies 50\leq f(10) \leq 62$ but we know $f(10)=3\cdot f(5)$ $\implies 17\leq f(5) \leq 20$ $\implies f(15) \geq 7\cdot 17 = 119$ but $f(16) = f(2)\cdot f(2)\cdot f(2)\cdot f(2)=81$ contradiction. If $f(2)>3$ then $f(8)>64$ but $f(9) = 49$ contradiction. CASE 2 is similar to CASE 1. you know how to do it im too lazyyy and after proving all this we realise that the minimum value of $f(3)=9$

I remember that it have been in a Chinese competition. we claim that all the function satisfied the given condition is $f(x)=x^a$ where a is a real number.

But this is a function from the positive integers to the positive integers.

This gives that $a$ is a positive integer, which solves the problem!

Does it? I mean, we know that multiplicative functions defined on the reals are of the form $f(x) = x^a$, but does this also hold for multiplicative functions defined on the positive integers?

No, the original problem (the one from China) was also from $\mathbb{N}$ to some set ($\mathbb{R}$ or so), and had also the condition that the function is increasing.

Oh, I see, didn't know that the fact that $f$ is increasing changes a few things indeed.

Btw, here it is: http://www.mathlinks.ro/Forum/viewtopic.php?t=55870

So observe that $f(x)=x^2$ is a solution.We will prove that $f(3)\neq7,8$ For that we see that f(1)=1 and $f(2)\ge 3$.Also $f(16)<f(18)$ so $f(2)<4$ so $f(2)=3$.But also $f(27)<f(32)$ which is equivalent with $343,512<243$ which is a cotradiction.

Since strictly increasing $\ln f(e^x)=cx$. Hence $f(3)=3^k$, so minimum is $\boxed 9$.

ZETA_in_olympiad wrote: Since strictly increasing $\ln f(e^x)=cx$. Hence $f(3)=3^k$, so minimum is $\boxed 9$. I slightly doubt that that solution works, because although $g(x) = \ln(f(e^x))$ does indeed satisfy Cauchy's Functional Equation, it can only take in inputs satisfy the property that $e^x$ is an integer, that is to say that the domain of $g(x)$ is only a specific subset of the real numbers. In which case it's unclear whether the linearization to Cauchy's FE still works. Please correct me if your reasoning still works. Anyway, by using a limit argument you can prove that $f(2) = f(3)^{\log_3 2}$ which is an integer iff $f(3)$ is a power of $3$, in which case the minimum is $9$. This is a little overkill but it generalizes nicely. Actually, I guess you need to prove that $9$ is attainable. But that’s clear from the function $f(x) = x^2$.