So, I've been thinking about trying to prove this statement -
If $n=n_1n_2$ and $n_1$ and $n_2$ are relatively prime integers greater than 2, prove both $φ(n_1)$ and $φ(n_2)$ divide $φ(n)$.
In practice, this is obvious - for instance, when $n=21$, $n_1=3$, $n_2=7$, $φ(n_1)=φ(3)=2$ and $φ(n_2)=φ(7)=6$, both of which divide $φ(n)/2=φ(21)/2=12/2=6$.
How do I set about proving this though? Since I know that $n_1$ and $n_2$ are coprime, I can write a linear combination of the two such that $kn_1 + jn_2=1$ for some integers k and j. But otherwise, I'm lost on how to bring the phi function into the equality. Hints would be great!
If you prove the following you'll have both a nice way of calculating and working with Euler function and a proof of what you want.
Suppose
$$n=\prod_{k=1}^mp_k^{a_k}\;,\;\;p_i\;\;\text{primes}\;,\;\;a_i\in\Bbb N$$
then
$$\varphi(n)=n\prod_{k=1}^m\left(1-\frac1{p_k}\right)$$