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Proof of a formula involving Euler's totient function.

For positive integers $m$ and $n$ where $d=gcd(m,n)$, show that $$\phi(mn) = \phi(m)\phi(n)\frac{d}{\phi(d)}$$

This is just the generalization of the multiplicative property of phi function.I have tried to solve this in the same way as the proof of multiplicative property but couldn't get far.Please help.

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Saurabh
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1 Answers1

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Suppose $$d=q_1^{\gamma_1}\cdot\ldots\cdot q_k^{\gamma_k}\,,\,m=d\cdot p_1^{\alpha_1}\cdot\ldots\cdot p_r^{\alpha_r}\,,\,n=d\cdot t_1^{\beta_1}\cdot\ldots\cdot t_s^{\beta_s}$$with $\,p_i\,,\,q_i\,,\,t_i\,$ primes , $\,\alpha_i\,,\,\beta_i\,,\,\gamma_i\in\mathbb{N}\,$ . Then, we have$$\varphi(mn)=mn\prod_{i=1}^r\left(1-\frac{1}{p_i}\right)\prod_{j=1}^s\left(1-\frac{1}{t_j}\right)\prod_{l=1}^k\left(1-\frac{1}{q_l}\right)$$$$\varphi(m)=m\prod_{i=1}^r\left(1-\frac{1}{p_i}\right)\,,\,\varphi(n)=n\prod_{j=1}^s\left(1-\frac{1}{t_j}\right)$$$$\varphi(d)=d\prod_{l=1}^k\left(1-\frac{1}{q_l}\right). $$The wanted equality follows at once.

amWhy
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DonAntonio
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    Shouldn't this is correct $$\varphi(m)=m\prod_{i=1}^r\left(1-\frac{1}{p_i}\right) \prod_{l=1}^k\left(1-\frac{1}{q_l}\right)$$ same for $\varphi(n)$ – Saurabh Jun 17 '12 at 10:40
  • Of course, @Saurabh. Thank you, I'll edit it. – DonAntonio Jun 17 '12 at 10:45
  • @DonAntonio Have edited the answer? Please answer correctly for me. – user1942348 Apr 25 '15 at 15:07
  • This is not correct. See https://math.stackexchange.com/questions/114841/proof-of-a-formula-involving-eulers-totient-function-varphi-mn-varphi?noredirect=1&lq=1 – Algebear Sep 16 '18 at 20:17
  • This is not yet fixed. It is very confusing. But I think once you fix it based on @Saurabh, it is a simpler proof that the linked one. – Souverain Premier Oct 15 '21 at 13:48