The Sum of Positive Divisors of an Integer

# The Sum of Positive Divisors of an Integer

Recall from The Number of Positive Divisors of an Integer page that if $n \in \mathbb{Z}$ then $d(n)$ denotes the number of positive divisors of $n$ which is given explicitly by:

(1)\begin{align} \quad d(n) = \sum_{d \mid n}_{d > 0} 1 \end{align}

We will now introduce another very important function.

Definition: Let $n \in \mathbb{Z}$. The function $\sigma : \mathbb{Z} \to \mathbb{N}$ defined by $\displaystyle{\sigma(n) = \sum_{d \mid n}_{d > 0} d}$ is the Sum of Positive Divisors Function. |

*Once again we define $\sigma(0) = 0$ and it is not hard to see that $\sigma(n) = \sigma(-n)$ for all $n \in \mathbb{Z}$ so we will often times only make mention of $\sigma(n)$ for $n \in \mathbb{N}$.*

For example, if $n = 18$ then the positive divisors for $n$ are $1$, $2$, $3$, $6$, $9$, and $18$. Hence:

(2)\begin{align} \quad \sigma(18) = 1 + 2 + 3 + 6 + 9 + 18 = 39 \end{align}

We will now look at some important results regarding this function.

Proposition 1: Let $p \in \mathbb{N}$ be a prime number. Then $\sigma (p) = 1 + p$. |

**Proof:**If $p$ is prime then the only divisors of $p$ and $1$ and $p$, so $\sigma (p) = 1 + p$. $\blacksquare$

Proposition 2: Let $p \in \mathbb{N}$ be a prime number. Then $\sigma (p^2) = 1 + p + p^2$. |

**Proof:**If $p$ is prime, then the divisors of $p^2$ are $1$, $p$, and $p^2$. So $\sigma (p^2) = 1 + p + p^2$. $\blacksquare$

Proposition 3: Let $p \in \mathbb{N}$ be a prime number and let $k \in \mathbb{N}$. Then $\displaystyle{\sigma (p^k) = 1 + p + p^2 + ... + p^k = \frac{p^{k+1} - 1}{p - 1}}$. |

**Proof:**Use Proposition 2 and apply induction.

- Note that $(1 + p + p^2 + ... + p^k)(p - 1) = p^{k+1} - 1$, so:

\begin{align} \quad \sigma (p^k) = 1 + p + p^2 + ... + p^k = \frac{p^{k+1} - 1}{p - 1} \quad \blacksquare \end{align}

Proposition 4: Let $p, q \in \mathbb{N}$ be prime numbers. Then $\sigma(pq) = \sigma(p) \sigma(q)$. |

**Proof:**The divisors of $pq$ are $1$, $p$, $q$, and $pq$, so:

\begin{align} \quad \sigma (pq) = 1 + p + q + pq \end{align}

- Additionally:

\begin{align} \quad \sigma(p) \sigma(q) = (1 + p)(1 + q) = 1 + p + q + pq \end{align}

- Thus $\sigma(pq) = \sigma(p) \sigma(q)$ as desired. $\blacksquare$