7

Suppose $m,n \in \Bbb N$, $k$ is product of all prime number such that divide $m,n$

How to prove that:

$$ \phi(mn)=\phi(m)\phi(n) \frac k{\phi(k)}$$

Mikasa
  • 67,942
agustin
  • 823
  • See also http://math.stackexchange.com/questions/119911/proving-formula-involving-eulers-totient-function. – lhf Jul 28 '16 at 11:03

4 Answers4

5

Note that $$\frac{\phi{(m)}}{m}=\prod_{p|m}(1-\frac{1}{p})$$

Thus

\begin{align} \frac{\phi{(mn)}}{mn}=\prod_{p|mn}(1-\frac{1}{p}) & =\frac{\prod_{p|m}(1-\frac{1}{p})\prod_{p|n}(1-\frac{1}{p})}{\prod_{p|m, p|n}(1-\frac{1}{p})} \\ &=\frac{\prod_{p|m}(1-\frac{1}{p})\prod_{p|n}(1-\frac{1}{p})}{\prod_{p|k}(1-\frac{1}{p})} \\ & =\frac{\phi{(m)}}{m}\frac{\phi{(n)}}{n}\frac{k}{\phi{(k)}} \end{align}

Multiplying by $mn$ gives the desired equality.

Ivan Loh
  • 17,115
  • 2
    why $ \prod_{p|mn}(1-\frac{1}{p})$ =$\frac{\prod_{p|m}(1-\frac{1}{p})\prod_{p|n}(1-\frac{1}{p})}{\prod_{p|m, p|n}(1-\frac{1}{p})} $? – agustin Mar 24 '13 at 05:22
  • 1
    Any $p$ which divides exactly one of $m, n$ is counted exactly once in the numerator, while any $p$ which divides both is double counted, so we have to divide by those, as shown in the denominator. – Ivan Loh Mar 24 '13 at 05:25
2

If $m$ and $n$ are coprime, then it is easy to show that $\phi(mn)=\phi(m)\phi(n)$. So we assume that $m=p^a$and $n=p^b$. Then $\phi(mn)=\phi(p^{a+b})=p^{a+b-1}(p-1)=p^{a-1}(p-1)p^{b-1}(p-1)\frac{p}{p-1}$. So the result follows.

P.S. When $m$ and $n$ are relatively prime, if $k$ is coprime to $mn$ then $k$ is prime to $m$ and $n$. Conversely, if $k_1$ is prime to $m$ and $k_2$ is prime to $n$, then there is a number prime to $mn$ and $\equiv k_1\pmod m$ and $\equiv k_2\pmod n$. And hence $\phi(mn)=\phi(m)\phi(n)$.

Point to me any error that occurs, thanks.

awllower
  • 16,926
0

Every Prime divisor of $mn$ is either a prime divisor of $m$ or a prime divisor of $n$. Those primes which divide both $m$ and $n$ also divide $(m,n)$, i.e., they also divide the GCD of $m$ and $n$ since GCD is but a multiplication of common prime divisors(factors) with least power.
Therefore, \begin{align} \frac{\phi{(mn)}}{mn}=\prod_{p|mn}(1-\frac{1}{p}) & =\frac{\prod_{p|m}(1-\frac{1}{p})\prod_{p|n}(1-\frac{1}{p})}{\prod_{p|m, p|n}(1-\frac{1}{p})} \\ &=\frac{\prod_{p|m}(1-\frac{1}{p})\prod_{p|n}(1-\frac{1}{p})}{\prod_{p|k}(1-\frac{1}{p})} \\ & =\frac{\phi{(m)}}{m}\frac{\phi{(n)}}{n}\frac{k}{\phi{(k)}} \end{align}

0

The equation is true, with the same proof, for any multiplicative function whose $p$ component does not depend on the power of $p$ dividing a number: $f(pn) = f(p^k n)$ for all $k > 0$.

These functions occur in estimates of the density of primes in polynomial sequences. For one or more polynomials with integer coefficients, the fraction of values for which the polynomials are all relatively prime to $n$ is a multiplicative function with the same property as $\frac{\phi(n)}{n}$, which is the special case of one polynomial equal to $P(n)=n$.

zyx
  • 36,077