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Math Help - Sum of divisors

  1. #1
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    Sum of divisors

    [FONT=Times New Roman]I need to prove that for all positive integers n, N^2 is >=2
    Last edited by miz.perfect84; August 11th 2009 at 10:21 AM.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by miz.perfect84 View Post
    I need to prove that for all positive integers n, σ (2n) > 2σ (n), where σ (n) denotes the sum of the divisors of n. [Hint: put n = \color{red}2^r\color{black}m, where r > = 0 and m is odd, and consider the cases r = 0 and r >0.]
    Use the fact that if \gcd(m,n)=1, then \sigma(mn)=\sigma(m)\sigma(n). If n is odd, then \sigma(2n)=\sigma(2)\sigma(n)=3\sigma(n)>2\sigma(n  ). Otherwise, n=2^rm where r\ge1 and m is odd, so

    \sigma(2n)\ =\ \sigma(2^{r+1})\sigma(m)
    =\ (1+2+2^2+\cdots+2^{r+1})\sigma(m)
    >\ (2+2^2\cdots+2^{r+1})\sigma(m)
    =\ 2(1+2+\cdots+2^r)\sigma(m)
    =\ 2\sigma(2^r)\sigma(m)
    =\ 2\sigma(2^rm)
    =\ 2\sigma(n)
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  3. #3
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    thanks

    that made it perfectly clear, thank you
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