Prove that if $\phi(n)\mid n$ then there is $ a \in \mathbb Z^+$ so that $a\ \times\prod_{i\ =\ 1}^{k}{{(p}_i-1)}\ =\prod_{i\ =\ 1}^{k}p_i$

Question

Given that $ n \in \mathbb Z^+$, $p_1,p_2,...p_k$ are its prime divisors I need to prove that if $\phi(n)\mid n$ then there is an $a \in \mathbb Z^+$ such that

$$a\ \times\prod_{i\ =\ 1}^{k}{{(p}_i-1)}\ =\prod_{i\ =\ 1}^{k}p_i.$$

Answer 1

Use that $\phi(n) = n\prod_{p_i \mid n} 1 - \frac{1}{p_i}$, where the product ranges over all the prime numbers. If you don’t already know this fact, prove it using that $\phi(ab) = \phi(a)\phi(b)$ when $(a,b) = 1$, $\phi(p^m) = p^m\left( 1 - \frac 1p\right)$ and the factorization of $n$ in primes.

If you know this fact then the problem becomes trivial , since there’s an $a$ such that $a \phi(n) = n$ by your hypothesis.