# Thread: Sum of positive divisor

1. ## Sum of positive divisor

Need help on the following:

Show a positive integer $n$ is composite iff $\sigma(n) > n + \sqrt{n}$.
$\sigma(n)$ denote the sum of positive divisor of n.

Thank you for your help.

2. Originally Posted by kleenex
Need help on the following:

Show a positive integer $n$ is composite iff $\sigma(n) > n + \sqrt{n}$.
$\sigma(n)$ denote the sum of positive divisor of n.

Thank you for your help.
Say $n>1$.

$n$ is prime if and only if $\sigma (n) = n+1$. Thus, $n$ is compositive if and only if $\sigma (n) > n+1 \mbox{ or }\sigma (n) < n+1$. But we cannot have $\simga(n) < n+1$ because $\sigma (n) > n+1$ (since $n$ and $1$ are always divisors). Thus, $n$ is compositive if and only if $\sigma (n) > n+1$. Thus, if $\sigma (n) > n+\sqrt{n} >n+1$ and so it is compositive.

3. Thank you,

I was thinking:
Let $d|n$ where $1 so $1. If $d \leq \sqrt n$ then $n/d \geq \sqrt n$ but then I don't know what next. But your way is easier than my.

On the other hand you got syntax error.

4. OK, to complete your proof you have to show that if a number $n$ doesn't have a prime divisor less or equal than $\sqrt[]{n}$ it's prime

Indeed, suppose it doesn't have have a prime divisor less or equal than $\sqrt[]{n}$, then n has at least two prime divisors $p_1,p_2$ (if it's composite) such that $p_1\cdot{p_2}\leq{n}$.

But $p_1,p_2>{\sqrt[]{n}}$ so the product $p_1\cdot{p_2}>n$ and this is absurd

So every composite number $n$ has at least one prime divisor $p$ such that $p\leq{\sqrt[]{n}}$

But you can only achieve the equality if $n$ is a perfect square, and in that case $\sigma(n)\geq{n+\sqrt[]{n}+1}>n+\sqrt[]{n}$ so the inequality holds

Otherwise it's not a perfect square and it has a prime divisor $p$ such that $p<\sqrt[]{n}$
So $\frac{n}{p}>\sqrt[]{n}$ is a divisor of $n$ and therefore $\sigma(n)>n+\sqrt[]{n}$