Sigma Function of Prime Number

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Theorem

Let $n$ be a positive integer.

Let $\map \sigma n$ be the sigma function of $n$.


Then $\map \sigma n = n + 1$ if and only if $n$ 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 $\map \sigma n$, 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 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:

$\map \sigma n \ge n + 1 + r + s > n + 1$

$\blacksquare$


Also see


Sources