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It is well known that Euler's totient satisfies $ \phi(mn) = \phi(m) \phi(n) \frac{d}{\phi(d)}, $ where $d = \gcd(m,n)$. By setting $ f(x)=\frac{\phi(x)}{x} $ this can be written as $ f(mn)f(d) = f(m)f(n) $

Have the functions that satisfy this equation been studied? They are generalized multiplicative functions.

Another generalization might be $ f(l)f(d) = f(m)f(n) $ where $l=\text{lcm}(m,n)$. The identity function satisfies this equation.

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    One complication is that $\frac{\phi(d)}{d}$ takes rational values.2012-03-13

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Let $g(n)=f(n)/f(1)$; then $g(mn)g(d)=f(mn)f(d)/(f(1))^2=f(m)f(n)/(f(1))^2=g(m)g(n)$ so $g$ has the same property and $g(1)=1$. So let's assume $f(1)=1$.

Now if $p$ is prime then $f(p^{n+1})f(p)=f(p^n)f(p)$ so $f(p^n)=f(p)$ for all $n\ge1$. Then $f(n)=\prod_{p\mid n}f(p)$ for all $n$. That is, $f$ depends only on the squarefree kernel of $n$.

I'm not sure how much there is to study about such functions.

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    Ah, [wikipedia](http://en.wikipedia.org/wiki/Multiplicative_function#Properties) says that *every* multiplicative function satisfies $f(l)f(d) = f(m)f(n)$. Of course. Silly me.2012-03-23