The Sum of Positive Divisors of an Integer

In number theory, σ(n) is the function which tells us the sum of all positive divisors for some integer n. For example, the positive divisors of n = 6 is 1, 2, 3, and 6. Thus the sum is 1 + 2 + 3 + 6 = 12. Hence σ(6) = 12. More formally, we define σ(n) to be:

(1)

\begin{align} \sigma(n) = \sum_{d \mid n} d \end{align}

Lemma 1: If p is a prime, then σ(p) = 1 + p.

• Proof: By the definition, a prime has only two positive divisors, namely 1 and p. Thus σ(p) = 1 + p.

Lemma 2: If p is prime, then σ(p²) = 1 + p + p²

• Proof: Once again, the only divisors of p² are 1, p, and p² since 1 | p², p | p², and p² | p². Thus σ(p²) = 1 + p + p².

Lemma 3: If p is a prime and e = 1, 2, 3, …, then σ(p^e) = 1 + p + p² + … + p^e

• Proof: Suppose we have a prime, p. Thus p^e has divisors 1 | p^e, p | p^e, p² | p^e …, p^e | p^e. Thus it follows that σ(p^e) = 1 + p + p² + … + p^e.

Lemma 4: Suppose p and q are primes, then σ(pq) = σ(p)σ(q).

• Proof: Since p and q are both primes, the only divisors of pq are 1, p, q, and pq. Thus it follows that:

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

• Let's factor out a q from the last two terms to obtain:

(3)

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

• And let's factor 1 + p out to obtain:

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

• But we know that σ(p) = 1 + p, and σ(q) = 1 + q. Thus σ(pq) = σ(p)σ(q).

Lemma 5: Suppose p and q are primes, then σ(pq²) = σ(p)σ(q²).

• Proof: Since p and q are both primes, the only divisors of pq² are 1, p, q, pq, q², and pq². Thus it follows that:

(5)

\begin{align} \sigma(pq^2) = 1 + p + q + pq + q^2 + pq^2 \end{align}

• We can factor the third/fourth and fifth/sixth terms to obtain:

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\begin{align} \sigma(pq^2) = 1 + p + q(1 + p) + q^2(1 + p) \end{align}

• Factoring out the 1 + p we obtain:

(7)

\begin{align} \sigma(pq^2) = (1 + p)(1 + q + q^2) \end{align}

• But σ(p) = 1 + p, and σ(q²) = 1 + q + q². Thus it follows that σ(pq²) = σ(p)σ(q).