Hi everybody. Show that for some positive integer $\displaystyle n$ is composite iff $\displaystyle \sigma(n)>n+\sqrt{n}$. Thanks.
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Hi Fernando, Try thinking about what $\displaystyle \sigma(n)$ is when $\displaystyle n$ is not composite (i.e. it is either prime or 1). - Hollywood
Thanks, the proof is by contradiction, is it not? Thanks for your suggestion.
Originally Posted by Fernando Hi everybody. Show that for some positive integer $\displaystyle n$ is composite iff $\displaystyle \sigma(n)>n+\sqrt{n}$. Thanks. $\displaystyle \sigma(n) =\begin{cases}n+1, & \text{if }n\text{ is prime} \\\displaystyle\prod_{i=1}^r \left(1+p_i+p^2_i+\ldots p^{a_i}_i\right), & \text{if }n\text{ is composite}\end{cases}$ where r is the number of distinct prime factors of n .
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