Sigma 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

From Rule of Transposition, we may replace the only if statement by its contrapositive.

Therefore, the following suffices:

Implication

Suppose $n$ is a 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 $n + 1$.

$\Box$

Contrapositive Implication

Suppose $n$ is not a prime.

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

As $n$ is composite, $\exists r, s \in \N: r, s > 1: r s = n$.

Trivially, both $r$ and $s$ are divisors of $n$.

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

$\blacksquare$