# Proof of a formula involving Euler's totient function: $\varphi (mn) = \varphi (m) \varphi (n) \cdot \frac{d}{\varphi (d)}$

The third formula on the wikipedia page for the Totient function states that $$\varphi (mn) = \varphi (m) \varphi (n) \cdot \dfrac{d}{\varphi (d)}$$ where $d = \gcd(m,n)$.

How is this claim justified?

Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $\varphi$ is multiplicative?

• There might be a direct proof, but of course if you show that $\varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $\varphi(p^a) = p^a - p^{a-1}$, then you get your result. Feb 29, 2012 at 15:18
• It'd be nice to relate this formula with the natural mapping $U_{mn}\to U_m \times U_n$ by proving that the kernel has size $d$ and the image has index $\phi(d)$.
– lhf
Mar 13, 2012 at 2:22
– lhf
Mar 14, 2012 at 12:07

You can write $$\varphi(n)$$ as a product $$\varphi(n) = n \prod\limits_{p \mid n} \left( 1 - \frac 1p \right)$$ over primes. Using this identity, we have

$$\varphi(mn) = mn \prod_{p \mid mn} \left( 1 - \frac 1p \right) = mn \frac{\prod_{p \mid m} \left( 1 - \frac 1p \right) \prod_{p \mid n} \left( 1 - \frac 1p \right)}{\prod_{p \mid d} \left( 1 - \frac 1p \right)} = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}$$

• This should probably be restricted to prime $p$. Apr 25, 2014 at 19:55

Hint $$\$$ A multiplicative function $$\rm\:f(n)\:$$ satisfies said identity if for all primes $$\rm\:p\:$$

$$\ \ \ \ \rm\ j\le k\ \Rightarrow\ \ f(p^{j+k}) = \frac{f(p^j)\: f(p^k)\: p^j}{f(p^j)}\ =\ p^j f(p^k)$$

Indeed we have $$\rm\ \ \phi(p^{j+k})\ =\ p^{j+k}-p^{j+k-1}\ =\ p^j (p^k-p^{k-1})\ =\ p^j \phi(p^k)$$

• Thanks, Bill. I'll try to wrap my head around that. It seems useful. Feb 29, 2012 at 16:44

Just as an additional note to the identity provided by @Cardboard Box.

$$\prod_{p|mn} (1 - \dfrac{1}{p}) = \frac{\prod_{p|m} (1 - \dfrac{1}{p}) \prod_{p|n} (1 - \dfrac{1}{p})}{\prod_{p|d} (1 - \dfrac{1}{p})}$$

Why does this work? Consider the prime factorization of $$m,n$$, and that we are multiplying all $$p$$ such that $$p|mn$$. But, on first sight, this is the same as multiplying all $$p$$ such that $$p|n$$ together with all $$p$$ such that $$p|m$$. But what if $$m$$ and $$n$$ share a prime factor?

This means that we will multiply that prime factor (call it $$p_c$$) twice, although of course it can only appear once in the prime factorization of $$mn$$, albeit with a greater power associated.

Thus, we need to account for this shared factor, or even shared factors in case there is more than 1 shared prime factor.

To do this, divide the whole thing by prime factors of the $$gcd(m,n)$$. Thus, we will be "removing" the factors which we multiplied twice.

Furthermore, to go from $$mn \frac{\prod_{p|m} (1 - \dfrac{1}{p}) \prod_{p|n} (1 - \dfrac{1}{p})}{\prod_{p|d} (1 - \dfrac{1}{p})} = \phi(m) \phi(n) \dfrac{d}{\phi(d)}$$

consider multiplying the left side by $$\dfrac{d}{d}$$. Thus you can group together the $$d$$ with $$\prod_{p|d} (1 - \dfrac{1}{p})$$ in the denominator to get $$\phi(d)$$.

Hope this makes it easier to follow the top answer.