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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- Oct 12th 2008, 01:20 PMmndi1105Sum 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) - Oct 12th 2008, 02:36 PMThePerfectHacker
If $\displaystyle n > 1$ is composite and not a square then the factors of $\displaystyle n$ include $\displaystyle 1,a,b,n$ where $\displaystyle ab = n$. This means that $\displaystyle \sigma(n) \geq 1 + a + b + n \geq n + 1 + 2\sqrt{ab} = n + 1 + 2\sqrt{n} > n + \sqrt{n}$.

If $\displaystyle n>1$ is composite and a square then the factors of $\displaystyle n$ include $\displaystyle 1,\sqrt{n},n$.

This means that $\displaystyle \sigma (n) \geq n + \sqrt{n} + 1 > n + \sqrt{n}$.

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