Totient function; if $\phi(ap)=\phi(a)\phi(p)$ and $p$ is prime, then $p\nmid a$

Question

If p is a prime and $\phi(ap)=\phi(a)\phi(p)$ can one conclude that a and p are relatively prime? I need to show that p does not divide a, but I'm not sure if $\phi(ap)=\phi(a)\phi(p)$ is enough to show that.

Answer 1

Yes, $\varphi(ap) = \varphi(a)\varphi(p)$ is enough to conclude that $a$ and $p$ are coprime. For $p$ prime, if $p$ divided $a$, we would have $\varphi(ap) = \varphi(a)\cdot p > \varphi(a)\varphi(p)$. More generally, you have

$$\varphi(n) = n\cdot \prod_{p\mid n}\left(1 - \frac1p\right),$$

and if $m$ and $n$ have a common prime factor $p$, the factor $\left( 1 - \frac1p\right)$ appears only once in $\varphi(mn)$, but twice in $\varphi(m)\varphi(n)$, so it is not necessary that one factor is prime to conclude the coprimality from $\varphi(mn) = \varphi(m)\varphi(n)$.

Answer 2

Suppose that $a=p^sb$, with $p,b$ co-prime and $s>0$. Then: $$\phi(ap)=\phi(p^{s+1}b)=p^s(p-1)\phi(b)$$ and: $$\phi(a)\phi(p)=p^{s-1}(p-1)\phi(b)(p-1).$$ If $\phi(ap)=\phi(a)\phi(p)$, then $p=p-1$, a contradiction.