Divisor Sum of Prime Number

Theorem
Let $n$ be a positive integer.

Let $\sigma \left({n}\right)$ be the sigma function of $n$.

Then $\sigma \left({n}\right) = n + 1$ iff $n$ is prime.

Proof

 * Suppose $n$ is prime.

By definition, the only positive divisors of $n$ are $1$ and $n$ itself.

Therefore $\sigma \left({n}\right)$, defined as the sum of the divisors of $n$, equals $n + 1$.


 * Suppose $n$ is not prime.

From One Divides All Integers and Every Integer Divides Itself, both $1$ and $n$ are divisors of $n$.

As $n$ is composite, $\exists r, s \in \N: r, s > 1: r s = n$ and so both $r$ and $s$ are divisors of $n$.

Hence $\sigma \left({n}\right) \ge n + 1 + r + s \ne n + 1$.