f Function

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

Why does

f(p)=f(2p) where p is an odd number?
Why does
2f(e)=f(2e) where e is an even number?

Why is f(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 f 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 f function we get:
new7

So f(n)=f(2n) (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 ł 2

f(n)=f(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 ł 1 and
gcd
(p^k,m)=1

.

The f function is a multiplicative function meaning, f(u v)=f(u)f(v) if
gcd
(u,v)=1

.

So we have:

f(n)=f(p^k)f(m)=p^k-1f(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.