For $a,b$ coprime, it is known that $\phi(ab)=\phi(a)\phi(b)$. But is there a connection between $\phi(ab)$ and $\phi(a),\phi(b)$ if they are not coprime?
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1See http://math.stackexchange.com/questions/114841/proofofaformulainvolvingeulerstotientfunction/114847 and http://math.stackexchange.com/questions/119911/provingformulainvolvingeulerstotientfunction. – lhf Oct 22 '13 at 16:58
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Note that Euler's product formula $$\phi(n) = n\prod_{p\mid n}\left(1\frac1p\right)$$ implies $$\begin{align} \phi(a\cdot b) &= ab\prod_{p\mid ab}\left(1\frac1p\right) \\ &= ab\frac{\prod_{p\mid a}\left(1\frac1p\right)\prod_{p\mid b}\left(1\frac1p\right)}{\prod_{p\mid\gcd(a,b)}\left(1\frac1p\right)} \cdot\frac{\gcd(a,b)}{\gcd(a,b)} \\ &= \phi(a)\phi(b)\frac{\gcd(a,b)}{\phi(\gcd(a,b))} \end{align}$$
Or to obtain a more symmetrical expression:
$$\phi(ab)\phi(\gcd(a,b)) = \phi(a)\phi(b)\gcd(a,b)$$
or even more symmetrical (courtesy of lhf):
$$\frac{\phi(ab)}{ab}\cdot\frac{\phi(d)}{d} = \frac{\phi(a)}a\cdot\frac{\phi(b)}b\quad\text{where}\ d = \gcd(a,b)$$
Tobias Kienzler
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1A very symmetrical expression is $$\def\F#1{\frac{\phi(#1)}{#1}} \F{ab} \F{d} = \F{a} \F{b}$$where $d=\gcd(a,b)$. – lhf Oct 23 '13 at 17:31


Note to self: The result somehow reminds me of Bayes' Theorem, though that might just be coincidence... – Tobias Kienzler Apr 12 '18 at 07:08