What are we supposed to do?

nayel wrote: What are we supposed to do? Whoops, sorry

WakeUp wrote: The real-valued function $f$ is defined for all positive integers. For any integers $a>1, b>1$ with $d=\gcd (a, b)$, we have \[f(ab)=f(d)\left(f\left(\frac{a}{d}\right)+f\left(\frac{b}{d}\right)\right) \]Determine all possible values of $f(2001)$. Let $P(x,y)$ be the assertion $f(xy)=f(\gcd(x,y))\left(f\left(\frac x{\gcd(x,y)}\right)+f\left(\frac y{\gcd(x,y)}\right)\right)$ Let $f(1)=a$ If $a=0$, then $P(3,667)$ $\implies$ $f(2001)=a(f(3)+f(667))=0$ If $a\ne 0$. Let $x>1$ : $P(x,x)$ $\implies$ $f(x^2)=f(x)(f(1)+f(1))=2af(x)$ $P(x^2,x)$ $\implies$ $f(x^3)=f(x)(f(x)+f(1))=f(x)^2+af(x)$ $P(x^3,x)$ $\implies$ $f(x^4)=f(x)(f(x^2)+f(1))=2af(x)^2+af(x)$ $P(x^2,x^2)$ $\implies$ $f(x^4)=f(x^2)(f(1)+f(1))=4a^2f(x)$ and so $2af(x)^2+af(x)=4a^2f(x)$ $\iff$ $f(x)(2f(x)+1-4a)=0$ and so : $\forall x$, either $f(x)=0$, either $f(x)=2a-\frac 12$ If $f(x)\ne 0$, we get $f(x)=2a-\frac 12$ and then either $f(x^2)=0$, either $f(x^2)=2a-\frac 12$ and so either $2af(x)=0$, either $2af(x)=2a-\frac 12$ and so $2af(x)=2a-\frac 12$ and so $(2a-1)(2a-\frac 12)=0$ So $a=\frac 12$ and $f(x)=f(x^2)=\frac 12$ And since the solution $f(x)=\frac 12$ $\forall x$ is indeed a solution, we get the result : $\boxed{f(2001)\in\{0,\frac 12\}}$ (and both are possible) [mod edit: also posted here and solved by pco.]