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 Integer Divisor Results, 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 $$\sigma \left({n}\right) \ge n + 1 + r + s \ne n + 1$$.