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Math Help - Sum of Positive Divisors Function

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    Sum of Positive Divisors Function

    Let sigma(n) be the sum of the positive divisors of n.

    Let n be an integer with n>0. Prove that n is composite if and only if sigma(n) > n + sqrt(n)
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    Quote Originally Posted by mndi1105 View Post
    Let n be an integer with n>0. Prove that n is composite if and only if sigma(n) > n + sqrt(n)
    If n > 1 is composite and not a square then the factors of n include 1,a,b,n where ab = n. This means that \sigma(n) \geq 1 + a + b + n \geq n + 1 + 2\sqrt{ab} = n + 1 + 2\sqrt{n} > n + \sqrt{n}.
    If n>1 is composite and a square then the factors of n include 1,\sqrt{n},n.
    This means that \sigma (n) \geq n + \sqrt{n} + 1 > n + \sqrt{n}.

    However, if n is not composite i.e. a prime then \sigma (n) = n + 1 < n + \sqrt{n}.
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