Prove $\phi(m)\phi(n)=\phi((m,n))\phi([m,n])$

Asked Feb 08 '23 at 11:02

Active Feb 08 '23 at 11:22

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Let $l=[m,n]$ be the lowest common multiple of $m$ and $n$ and let $d=(m,n)$ be the greatest common divisor of $m$ and $n$.

Prove $\phi(m)\phi(n)=\phi(l)\phi(d)$.

We know $mn=dl$.
We also know $\phi(mn)=\phi(m)\phi(n)\frac{d}{\phi(d)}$.

It is clear that $(d,l)=d$. Now appliying the $\phi$ to 1. and using 2. we get, $$\phi(mn)=\phi(dl)$$ $$\phi(m)\phi(n)\frac{d}{\phi(d)}=\phi(d)\phi(l)\frac{d}{\phi(d)}.$$ $\frac{d}{\phi(d)}$ cancels on both sides, thus the statement is proven.

Is this correct?

edited Feb 08 '23 at 11:10

metamorphy

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asked Feb 08 '23 at 11:02

Shean

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Does this answer your question? Prove $\varphi(a)\varphi(b)=\varphi(\gcd(a,b))\varphi(\operatorname{lcm}[a,b])$ - found using an Approach0 search. – John Omielan Feb 08 '23 at 12:48
It provides a good proof, but I am still wondering if my solution is correct. – Shean Feb 08 '23 at 15:56
1

Yes, your solution is also correct, as long as you can use the $\phi(mn)=\phi(m)\phi(n)\frac{d}{\phi(d)}$ result without proving it (FYI, this relation is listed in the third bullet point, and the result you're asked to prove is the fourth bullet point, in the Other formulae section of Wikipedia's "Euler's totient function" article). – John Omielan Feb 08 '23 at 18:26

Prove $\phi(m)\phi(n)=\phi((m,n))\phi([m,n])$

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