Let $d=(m,n)$. Prove: $$φ (mn) = φ (m) φ (n)\frac{d}{φ (d)}$$ Can anyone help me with this proof?

2Do you know a nice formula for $\dfrac{\varphi(k)}{k}$? – Daniel Fischer Jan 10 '16 at 16:43

$\sum_{k=1}^n\frac{\varphi(k)}{k} = \sum_{k=1}^n\frac{\mu(k)}{k}\left\lfloor\frac{n}{k}\right\rfloor=\frac6{\pi^2}n+O\left((\log n)^{2/3}(\log\log n)^{4/3}\right) $ @DanielFischer – mod Jan 10 '16 at 16:46

I meant for one term, not for the sum. Can you compute $\varphi(k)$ from the prime factorisation of $k$? – Daniel Fischer Jan 10 '16 at 16:48

Do u mean this $\varphi(k) =k \prod_{p\mid k} \left(1\frac{1}{p}\right)$ ? @DanielFischer – mod Jan 10 '16 at 16:52

1Exactly that. Now use that for $k \in \{m,n,mn,d\}$ and look sharp. – Daniel Fischer Jan 10 '16 at 16:53

1@DanielFischer, please make it into an answer, so that I can upvote it. It's really inclusionexclusion on $\varphi(x)/x$. – Andreas Caranti Jan 10 '16 at 17:06
1 Answers
I use the fact that phifunction is multiplicative.
If a prime $p^{\alpha1}m$ and $p$ does not divide $n$ then it contributes $p^{\alpha1}(p1)$ to the left hand side. Also its obvious that it will "create" a factor of $p^{\alpha1}(p1)$ on the right hand side.
If a prime $p^{\alpha1}n$ and $p$ does not divide $m$ then it contributes $p^{\alpha1}(p1)$ to the left hand side. Also its obvious that it will "create" a factor of $p^{\alpha1}(p1)$ on the right hand side.
If a prime $q$ such that $q^{r}m$ and $q^{s}n$ then it contributes a factor $q^{r+s1}(q1)$ on the left and side. On the right hand side the $q$ will contribute a factor of $\frac{q^{r1}(q1)q^{s1}(q1)q^{min(m,n)}}{q^{min(m,n)1}(q1)}=q^{r+s1}(q1)$.
So, we are done.
We can rigorise this using canonical factorisation representation of the $m$ and $n$.
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