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If $m$ and $n$ be positive integers Prove that $\varphi(mn)= \varphi(m,n) \cdot \varphi[m,n]$ where [m, n] =l.c.m of $a$ and $b$ And (m, n)=g.c.d of $a$ and $b$

My approach $\varphi (mn)=\varphi ((m, n) [m, n])$ $\implies ({mn/[m, n]} ,[m, n])$ $=({mn/[m, n]},{mn/(m, n)}$ Am I in right direction

John757
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  • If $m=n=p$, a prime, then you are claiming that $\varphi(p^2)=\varphi(p)^2$. – lulu Jun 27 '18 at 10:19
  • Can you clarify your question? As stated, what you want to prove is not true. – lulu Jun 27 '18 at 10:26
  • what you claim is incorrect, a possible identity is provided in https://math.stackexchange.com/questions/2303291/eulers-totient-function-of-a-product-for-arbitrary-n-and-m – AlvinL Jun 27 '18 at 10:26
  • @lulu OK my approach could be wrong so I seeked help – John757 Jun 27 '18 at 10:28
  • It's not a problem with the approach. The claim you are trying to prove is false. – lulu Jun 27 '18 at 10:29
  • @lulu OK then then please give a hint what I need to prove – John757 Jun 27 '18 at 10:31
  • I have no idea what you need to prove! The claim you want to prove isn't true. I don't see any sensible way to modify it to get a true statement (though of course I might be missing something). Another commenter posted a link to a question with a formula in it...maybe that's something you are interested in? – lulu Jun 27 '18 at 10:34
  • @lulu the first paragraph is the question. I need to prove what is stated there – John757 Jun 27 '18 at 10:36
  • Sorry, we are going in circles. The statement in the first paragraph is simply not true. I gave you a counterexample. Please study the counterexample. – lulu Jun 27 '18 at 10:38

1 Answers1

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As mentioned in the comments, it is not true that $$ \varphi(mn)= \varphi(m,n) \varphi[m,n] $$ For instance, it fails for $m=n=p$, where $p$ is prime, because $\varphi(p^2)=p(p-1) \ne p^2 = \varphi(p)^2$.

The correct identity is $$ \varphi(mn) = \varphi(m) \varphi(n) \frac{d}{\varphi(d)} $$ where $d = \gcd(m,n)$. See a proof here.

This symmetric version of that identity looks much nicer $$ \frac{\varphi(mn)}{mn} \frac{\varphi(d)}{d} = \frac{\varphi(m)}{m} \frac{\varphi(n)}{n} $$

lhf
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