Euler phi at a product

If the positive greatest common divisor of the integers $a$ and $b$ is $d$, then

$$\phi (ab)=\frac{\phi (a)\phi (b)d}{\phi (d)}.$$

Proof.  Using the positive prime factors $p$, the right hand side of the asserted equation is

${\displaystyle \frac{d\cdot a{\prod }_{p\mid a}\frac{p-1}{p}\cdot b{\prod }_{p\mid b}\frac{p-1}{p}}{d{\prod }_{p\mid a,p\mid b}\frac{p-1}{p}}}$ $\mathrm{\hspace{0.33em}}={\displaystyle \frac{ab{\prod }_{p\mid a,p\nmid b}\frac{p-1}{p}\cdot {\prod }_{p\mid a,p\mid b}\frac{p-1}{p}\cdot {\prod }_{p\mid b,p\nmid a}\frac{p-1}{p}\cdot {\prod }_{p\mid b,p\mid a}\frac{p-1}{p}}{{\prod }_{p\mid a,p\mid b}\frac{p-1}{p}}}$

$\mathrm{\hspace{0.33em}}=ab{\displaystyle \prod _{p\mid a\vee p\mid b}}{\displaystyle \frac{p-1}{p}}=ab{\displaystyle \prod _{p\mid ab}}{\displaystyle \frac{p-1}{p}}=\phi (ab),$

Q.E.D.

Title	Euler phi at a product
Canonical name	EulerPhiAtAProduct
Date of creation	2014-02-18 14:02:24
Last modified on	2014-02-18 14:02:24
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11A25
Classification	msc 11-00
Related topic	EulerPhifunction
Related topic	DivisibilityByPrimeNumber