The third formula on the wikipedia page for the Totient function states that $$\varphi (mn) = \varphi (m) \varphi (n) \cdot \dfrac{d}{\varphi (d)} $$ where $d = \gcd(m,n)$.

How is this claim justified?

Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $\varphi$ is multiplicative?

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    $\begingroup$ There might be a direct proof, but of course if you show that $\varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $\varphi(p^a) = p^a - p^{a-1}$, then you get your result. $\endgroup$ – Joel Cohen Feb 29 '12 at 15:18
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    $\begingroup$ It'd be nice to relate this formula with the natural mapping $U_{mn}\to U_m \times U_n$ by proving that the kernel has size $d$ and the image has index $\phi(d)$. $\endgroup$ – lhf Mar 13 '12 at 2:22
  • $\begingroup$ See also math.stackexchange.com/questions/119911/…. (Thanks @Dane!) $\endgroup$ – lhf Mar 14 '12 at 12:07

You can write $\varphi(n) = n \prod_{p \mid n} \left( 1 - \frac 1p \right)$. Using this identity, we have

$$ \varphi(mn) = mn \prod_{p \mid mn} \left( 1 - \frac 1p \right) = mn \frac{\prod_{p \mid m} \left( 1 - \frac 1p \right) \prod_{p \mid n} \left( 1 - \frac 1p \right)}{\prod_{p \mid d} \left( 1 - \frac 1p \right)} = \varphi(m)\varphi(n) \frac{d}{\varphi(d)} $$

  • $\begingroup$ This should probably be restricted to prime $p$. $\endgroup$ – G. Bach Apr 25 '14 at 19:55

Hint $\ $ A multiplicative function $\rm\:f(n)\:$ satisfies said identity if for all primes $\rm\:p\:$

$$\rm\ j\le k\ \Rightarrow\ \ f(p^{j+k}) = \frac{f(p^j)\: f(p^k)\: p^j}{f(p^j)}\ =\ p^j f(p^k)$$

Indeed we have $\rm\ \ \phi(p^{j+k})\ =\ p^{j+k}-p^{j+k-1}\ =\ p^j (p^k-p^{k-1})\ =\ p^j \phi(p^k)$

  • $\begingroup$ Thanks, Bill. I'll try to wrap my head around that. It seems useful. $\endgroup$ – The Chaz 2.0 Feb 29 '12 at 16:44

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