The real-valued function $f$ is defined for all positive integers. For any integers $a>1, b>1$ with $d=\operatorname{gcd}(a, b)$, we have

$$
f(a b)=f(d) \cdot\left(f\left(\frac{a}{d}\right)+f\left(\frac{b}{d}\right)\right),
$$

Determine all possible values of $f(2001)$.
Answer: 0 and $\frac{1}{2}$.

Obviously the constant functions $f(n)=0$ and $f(n)=\frac{1}{2}$ provide solutions.

We show that there are no other solutions. Assume $f(2001) \neq 0$. Since $2001=3 \cdot 667$ and $\operatorname{gcd}(3,667)=1$, then

$$
f(2001)=f(1) \cdot(f(3)+f(667)),
$$

and $f(1) \neq 0$. Since $\operatorname{gcd}(2001,2001)=2001$ then

$$
f\left(2001^{2}\right)=f(2001)(2 \cdot f(1)) \neq 0 .
$$

Also $\operatorname{gcd}\left(2001,2001^{3}\right)=2001$, so

$$
f\left(2001^{4}\right)=f(2001) \cdot\left(f(1)+f\left(2001^{2}\right)\right)=f(1) f(2001)(1+2 f(2001))
$$

On the other hand, $\operatorname{gcd}\left(2001^{2}, 2001^{2}\right)=2001^{2}$ and

$$
f\left(2001^{4}\right)=f\left(2001^{2}\right) \cdot(f(1)+f(1))=2 f(1) f\left(2001^{2}\right)=4 f(1)^{2} f(2001) .
$$

So $4 f(1)=1+2 f(2001)$ and $f(2001)=2 f(1)-\frac{1}{2}$. Exactly the same
argument starting from $f\left(2001^{2}\right) \neq 0$ instead of $f(2001)$ shows that $f\left(2001^{2}\right)=2 f(1)-\frac{1}{2}$. So

$$
2 f(1)-\frac{1}{2}=2 f(1)\left(2 f(1)-\frac{1}{2}\right) \text {. }
$$

Since $2 f(1)-\frac{1}{2}=f(2001) \neq 0$, we have $f(1)=\frac{1}{2}$, which implies $f(2001)=2 f(1)-\frac{1}{2}=\frac{1}{2}$.