Let $d=(m,n)$. Prove: $$φ (mn) = φ (m) φ (n)\frac{d}{φ (d)}$$ Can anyone help me with this proof?
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2Do you know a nice formula for $\dfrac{\varphi(k)}{k}$? – Daniel Fischer Jan 10 '16 at 16:43
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$\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
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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
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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
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1Exactly that. Now use that for $k \in {m,n,mn,d}$ and look sharp. – Daniel Fischer Jan 10 '16 at 16:53
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1@DanielFischer, please make it into an answer, so that I can upvote it. It's really inclusion-exclusion on $\varphi(x)/x$. – Andreas Caranti Jan 10 '16 at 17:06
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I use the fact that phi-function is multiplicative.
If a prime $p^{\alpha-1}||m$ and $p$ does not divide $n$ then it contributes $p^{\alpha-1}(p-1)$ to the left hand side. Also its obvious that it will "create" a factor of $p^{\alpha-1}(p-1)$ on the right hand side.
If a prime $p^{\alpha-1}||n$ and $p$ does not divide $m$ then it contributes $p^{\alpha-1}(p-1)$ to the left hand side. Also its obvious that it will "create" a factor of $p^{\alpha-1}(p-1)$ on the right hand side.
If a prime $q$ such that $q^{r}||m$ and $q^{s}||n$ then it contributes a factor $q^{r+s-1}(q-1)$ on the left and side. On the right hand side the $q$ will contribute a factor of $\frac{q^{r-1}(q-1)q^{s-1}(q-1)q^{min(m,n)}}{q^{min(m,n)-1}(q-1)}=q^{r+s-1}(q-1)$.
So, we are done.
We can rigorise this using canonical factorisation representation of the $m$ and $n$.
Subham Jaiswal
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