Thread: 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.

Thanks in advance for your help.

If then there is one number ( ) which is not coprime to n, for instance the same n. Thus:
for n is not coprime to n

And obviously for n=1 the equality holds.


Try to do the second problem
Hint: Remember that (the sum of the positive divisors of n) is mulitplicative. This means that given two natural numbers such that we have

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.