17
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
28
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ This should probably be restricted to prime $p$. $\endgroup$ – G. Bach Apr 25 '14 at 19:55
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.