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Math Help - 2 problems I am having difficulties with

  1. #1
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    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.
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  2. #2
    Super Member PaulRS's Avatar
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    If n>1 then there is one number a ( 1\leq{a}\leq{n}) which is not coprime to n, for instance the same n. Thus: \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 \sigma(n) (the sum of the positive divisors of n) is mulitplicative. This means that given two natural numbers a,b such that (a,b)=1 we have \sigma(a\cdot{b})=\sigma(a)\cdot{\sigma(b)}
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  3. #3
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    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.
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