# 2 problems I am having difficulties with

• Apr 15th 2008, 02:28 PM
jonnyfive
2 problems I am having difficulties with
I have 2 problems that I need some help on if you guys would.

Show that phi(n)less than or equal to n, and phi(n)=n iff n=1.

and

For k> or equal to 2 show that if 2^k -1 is prime then n=(2^k-1)((2^k)-1) has o(n)=2n, where o(n) is sum of all positive divisors of n.

• Apr 15th 2008, 03:24 PM
PaulRS
If $\displaystyle n>1$ then there is one number $\displaystyle a$ ($\displaystyle 1\leq{a}\leq{n}$) which is not coprime to n, for instance the same n. Thus: $\displaystyle \phi(n)=\sum_{(a,n)=1;1\leq{a}\leq{n}}{1}\leq{\sum _{1\leq{a}<n}{1}}=n-1<n$
for n is not coprime to n

And obviously for n=1 the equality holds.

Try to do the second problem
Hint: Remember that $\displaystyle \sigma(n)$ (the sum of the positive divisors of n) is mulitplicative. This means that given two natural numbers $\displaystyle a,b$ such that $\displaystyle (a,b)=1$ we have $\displaystyle \sigma(a\cdot{b})=\sigma(a)\cdot{\sigma(b)}$ (Wink)
• Apr 15th 2008, 05:19 PM
jonnyfive
Thanks on the first one.

The second one, that is not phi, it is sigma.

and phi does stand for Euler's totient function.

Thanks again.