Number Theory and Euler phi-function

[pjc11]

Homework Statement

Let p be prime. Show that p ∤ n, where n is a positive integer, iff \phi(np) = (p-1)\phi(n).

Homework Equations

Theorem 1: If p is prime, then \phi(p) = p-1. Conversely, if p is a positive integer with \phi(p) = p-1, then p is prime.

Theorem 2: Let m and n be relatively prime positive numbers. Then \phi(mn) = \phi(m)\phi(n).

The Attempt at a Solution

Assume that p ∤ n.
By Theorem 2, \phi(np) = \phi(n)\phi(p).
By Theorem 1, \phi(np) = \phi(n)\cdot(p-1).
So, if p ∤ n, then \phi(np) = (p-1)\cdot\phi(n).

Now, assume that \phi(np) = (p-1)\phi(n).
By Theorem 1, \phi(np) = \phi(p)\phi(n), since p is prime.

...and, that's it. I can't use Theorem 2 to show that n and p are relatively prime, so I'm not entirely sure what to do next. p is prime, so n ∤ p, but I don't see anything to base a conclusion of p ∤ n.

Any help would be greatly appreciated!

 

[Dick]

If a prime p doesn't divide n then aren't p and n relatively prime? A common divisor of p and n must divide p. Think about it.