$ϕ$ Function

By Mary-Ellen Lynall on Saturday, February 02, 2002 - 01:43 pm:

Why does

ϕ (p) = ϕ (2 p)

where

p

is an odd number?
Why does

2 ϕ (e) = ϕ (2 e)

where

e

is an even number?

Why is $ϕ (n)$ always even when $n$ is bigger than 2?

By Yatir Halevi on Saturday, February 02, 2002 - 07:09 pm:

Your first question:
let n be a positive integer with the following prime factorization:
n=p₁ ^k₁ p₂ ^k₂ ...p_i ^k_i
for example: 10=2² 5²

The $ϕ$ function for this $n$ is defined as the following:
new6

If n is odd, so none of the primes (p₁ ,p₂ ,...,p_i ) are even. But 2n is even, and its prime factorization is:
n=2p₁ ^k₁ p₂ ^k₂ ...p_i ^k_i

Taking the

ϕ

function we get:
new7

ϕ (n) = ϕ (2 n)

(when

n

is odd)
Second question:
Trying doing it yourself, based on what I did just now. If you'll need more help, don't hesitate to ask.

Third question:
(This is taken from "Elementary Number Theory" by David M Burton)

Assume that

n

is a power of 2:

n = 2^{k}

with

k \geq 2

$ϕ (n) = ϕ (2^{k}) = 2^{k} (1 - (1 / 2)) = 2^{k - 1}$

an even integer. If $n$ isn't a power of 2, then it is divisible by an odd prime $p$ , we therefore can write $n$ this way: $n = p^{k} m$ $k \geq 1$ and $gcd (p^{k}, m) = 1$ .

The $ϕ$ function is a multiplicative function meaning, $ϕ (u v) = ϕ (u) ϕ (v)$ if $gcd (u, v) = 1$ .

So we have:

$ϕ (n) = ϕ (p^{k}) ϕ (m) = p^{k - 1} ϕ (m)$

which is even because 2 divides $(p - 1)$ , because $p$ is odd and therefore $(p - 1)$ is even.

And again if something is not clear enough....ask.

Yatir

By Mary-Ellen Lynall on Monday, February 04, 2002 - 04:55 pm:

Thankyou very much for your answers. They were very helpful.