Showing an equality of Euler Totient Function

Question

Show the following identity for Euler’s ϕ function:

$ϕ(m · n) = \frac{ϕ(m) · ϕ(n) · gcd(m, n)}{ϕ(gcd(m, n))}$

\begin{eqnarray*} \phi(m) = m \prod_{p \mid m } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \phi(n) = n \prod_{p \mid n } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \phi(m.n) = m.n \prod_{p \mid m.n } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \phi(gcd(m,n)) = gcd(m,n) \prod_{p \mid gcd(m,n) } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \$\frac{\phi(m).\phi(n).gcd(m,n)}{\phi(gcd(m,n))} = \frac{m \prod_{p \mid m } (1-\frac{1}{p}).n \prod_{p \mid n } (1-\frac{1}{p}).gcd(m,n)}{gcd(m,n) \prod_{p \mid gcd(m,n) } (1-\frac{1}{p})}=m.n \prod_{p \mid m.n } (1-\frac{1}{p})=\phi(m.n) \\ \end{eqnarray*}

Is this correct?

Answer 1

HINT: Express $m$ and $n$ in terms of their prime factorisations \begin{eqnarray*} m= \prod_{i=1}^{k} p_i^{ \alpha_i} \\ \end{eqnarray*} and use \begin{eqnarray*} \phi(r) = r \prod_{p \mid r } (1-\frac{1}{p}). \\ \end{eqnarray*}

Answer 2

Hint:
This is in fact correct, but there is a better way to do it than considering the product for the totient function.

Answer 3

Hint Use $$\frac{\phi(n)}{n}= \prod_{p |n } (1-\frac{1}{p})$$