Exercise (4.5).8

(a) Show that $𝜎 (2^n) = 2^{n+1} − 1$.

(b) Show that $𝜎 (p^n) = p^{n+1} − 1$ where $p$ is prime.

(a) The sigma function of a natural number is the sum of all the positive divisors of that number.

The divisors of $2^n$ are $1, 2, 2^2, 2^3, \ldots , 2^n$. And so

$$ \begin{align} \sigma(2^n) & = 1 + 2 + 2^2 + 2^3 \ldots 2^n  \\ \\ & = \frac{1-2^{n+1}}{1-2} \quad \text{geometric sum} \\ \\ & = 2^{n+1} - 1 \end{align}$$

Note: the formula for the sum of a geometric series is here (link).

(b) The assertion is wrong. Consider $p=3$ as a counter-example.

The divisors of $3^n$ are $1, 3, 3^2, 3^3, \ldots , 3^n$. And so

$$ \begin{align} \sigma(3^n) & = 1 + 3 + 3^2 + 3^3 \ldots 3^n  \\ \\ & = \frac{1-3^{n+1}}{1-3} \quad \text{geometric sum} \\ \\ & = \frac{3^{n+1} - 1}{2} \\ \\  & \ne 3^{n+1}-1 \end{align}$$

The correct expression is

$$ \sigma(p^n) = \frac{p^{n+1}-1}{p-1}$$

Exercise (4.5).7

Show that the following statement is false:

‘There is one perfect number for any given number of digits.’

The first few perfect numbers are: 6, 28, 496, 8128, 33550336 (src A000396).

This tells us there is no perfect number of five, six or seven digits.

Exercise (4.5).6

Show that the following is false:

‘An even number is an abundant number.’

We can show this with a counter-example.

Consider the even number 4. It has two proper factors, 1 and 2$. The sum is 3. which is less than 4, and so 4 is deficient, not abundant.

Another, even simpler example, is the even number 2, whose only proper factor is 1, and so the sum is 1, which is less than 2, leaving 2 as a deficient number.

Exercise (4.5).5

Prove that a prime number is a deficient number.

A number $n$ is deficient if the sum of its proper factors is less than $n$.

A prime number $p$ has only two factors, 1 and $p$ itself. It has only one proper factor, 1.

And so any prime is deficient because the sum of its proper factors is always 1, which is less than any prime.

Exercise (4.5).4

Determine $\sigma (500)$.

We use the multiplicativity of the sigma function, 

$$ \gcd(a,b)=1  \quad \implies \quad \sigma(a \times b) = \sigma(a) \times \sigma(b) $$

and Proposition (4.35), where $p$ is prime and $k$ is a positive integer,

$$ \sigma(p^k) = \frac{p^{k+1} - 1}{p - 1} $$

Noting $\gcd(2^2, 5^3)=1$, we have

$$\begin{align} \sigma(500) & = \sigma(2^2) \times \sigma(5^3) \\ \\  & =  \frac{2^3-1}{2-1} \times \frac{5^4-1}{5-1} \\ \\ & = 1092 \end{align} $$

Exercise (4.5).3

(a) Show that $n$ is an abundant number $\iff 𝜎 (n) > 2n$.

(b) Show that $n$ is a deficient number $\iff 𝜎 (n) < 2n$.

(c) Show that $n$ is a perfect number $\iff 𝜎 (n) = 2n$.

Characterise the numbers in question 2 into perfect, abundant, or deficient numbers.

(a) An abundant number $n$ is one where the sum of its proper factors $p(n)$ is greater than $n$.

Noting the definition $\sigma(n) = p(n) + n$, we have

$$ \text{abundant }n \iff p(n) > n \iff \sigma(n) > n + n  \iff \sigma(n) > 2n $$

(b) A deficient number $n$ is one where the sum of its proper factors $p(n)$ is less than $n$. 

Noting the definition $\sigma(n) = p(n) + n$, we have

$$ \text{deficient }n \iff p(n) < n \iff \sigma(n) < n + n  \iff \sigma(n) < 2n $$

(c) A perfect number $n$ is one where the sum of its proper factors $p(n)$ equals $n$. 

Noting the definition $\sigma(n) = p(n) + n$, we have

$$ \text{perfect }n \iff p(n) = n \iff \sigma(n) = n + n  \iff \sigma(n) = 2n $$

We characterise the numbers from exercise 2 as follows:

• 15 is a deficient number because $\sigma(15) = 24 < 30$
• 77 is a deficient number because $\sigma(77)=96 < 154$
• 171 is a deficient number because $\sigma(171) = 260 < 342$
• 200 is an abundant number because $\sigma(200)=465 > 400$

Exercise (4.5).2

Determine the following:

(a) $\sigma (15)$

(b) $\sigma (77)$

(c) $\sigma (171)$

(d) $\sigma (200)$

The sigma function of a natural number $n$ is the sum of its positive divisors, including 1 and $n$ itself.

(a) $\sigma (15) = 1 + 3 + 5 + 15 = 24$

(b) $\sigma (77) = 1 + 7 + 11 + 77 = 96$

(c) $\sigma (171) = 1 + 3 + 3^2 + 19 + (3\times19) + 171 = 260$

(d) $\sigma(200) = 1 + 2 + 2^2 + 2^3 + 5 + 5^2 + (2 \times 5) + (2^2 \times 5) + (2^3 \times 5) + (2 \times 5^2) + (2^2 \times 5^2) + 200 = 465 $