# Generalized multiplicative functions

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. – lhf Mar 13 '12 at 13:11

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 $n$.
Thanks! How about the second equation, when $f(l)f(d) = f(m)f(n)$? – lhf Mar 23 '12 at 10:17
The reduction to the case $f(1)=1$ still works, but you don't get any restriction on the value at prime powers. I don't know if anyone has looked at the set of functions with this property. – Gerry Myerson Mar 23 '12 at 11:10
Ah, wikipedia says that every multiplicative function satisfies $f(l)f(d) = f(m)f(n)$. Of course. Silly me. – lhf Mar 23 '12 at 11:56