If p divides n, then p-1 divides phi(n)

The Mathematician — Tue, 01 Nov 2022 13:00:48 +0000

Let $p$ be a prime number. If $p \mid n$ , then $p - 1 \mid \phi(n)$ .

Proof. We need to prove that if

p

divides

n

, then

p-1

divides the Euler’s totient function of

n

. Assume that

p \mid n

. Then

n = p^{\alpha}c

for some integer

c

and positive integer

\alpha

where

p

doesn’t divide

c

. Now since

p \nmid c

, this implies that

gcd(p^{\alpha},c) = 1

. Since the Euler’s totient function is multiplicative, we can apply that

\phi(p^{\alpha}c) = \phi(p^{\alpha})\phi(c)

. So together, we get

\begin{align*} \phi(n) &= \phi(p^{\alpha}c) \quad \quad \text{ since } n = p^{\alpha}c \\ &=\phi(p^m)\phi(c) \quad \text{ since } gcd(p^{\alpha},c) = 1 \\ &= p^{\alpha - 1}(p-1)\phi(c). \end{align*}

So since

\phi(n) = p^{\alpha - 1}(p-1)\phi(c)

, we have that

p - 1 \mid \phi(n)

p divides n Archives - Epsilonify