(a) Show that
$$ \sigma (p^3) = (p^2 + 1) (p + 1) $$
where $p$ is prime.
(b) Show that
$$ \sigma (p^5) = (p^2 − p + 1) (p^2 + p + 1) (p + 1) $$
where $p$ is prime.
(a) By definition of the sigma function
$$ \sigma (p^3) = 1 + p + p^2 + p^3 = (p^2+1)(p+1) $$
(b) By definition of the sigma function
$$ \sigma(p^5) = 1 + p + p^2 + p^3 + p^4 + p^5 = (p + 1) (p^2 - p + 1) (p^2 + p + 1) $$
Prove Proposition (4.37).
Proposition (4.37). Let the prime decomposition of a natural number $n$ be given by
$$ n = p_1^{k_1} \times p_2^{k_2} \times p_3^{k_3} \times \ldots \times p_m^{k_m} $$
where $p_j$’s are distinct primes. Then
$$ \begin{align} \sigma(n) & = \sigma(p_1^{k_1} \times p_2^{k_2} \times p_3^{k_3} \times \ldots \times p_m^{k_m}) \\ \\ & = \sigma(p_1^{k_1}) \times \sigma(p_2^{k_2}) \times \sigma(p_3^{k_3}) \times \ldots \times \sigma(p_m^{k_m}) \end{align} $$
We start by noting that the divisors of $n$ are of the form
$$ p_1^{a_1} \times p_2^{a_2} \times p_3^{a_3} \times \ldots \times p_m^{a_m} $$
where $0 \le a_i \le k_i$ where $1 \le i \le m$.
The sum of all such divisors is given by the following product. To explain this further, multiplying out the brackets results in summands that represent every possible divisor of the above form.
$$ \begin{align} \sigma(n) & = (\sum_{i=0}^{k_1} p_1^i) \times (\sum_{i=0}^{k_2} p_2^i) \times (\sum_{i=0}^{k_3} p_3^i) \times \ldots \times (\sum_{i=0}^{k_m} p_m^i) \\ \\ & = \sigma(p_1^{k_1}) \times \sigma(p_2^{k_2}) \times \sigma(p_3^{k_3}) \times \ldots \times \sigma(p_m^{k_m}) \end{align}$$
This is the desired result.
Note: the author's solution incorrectly applies Proposition (4.36).
Prove Proposition (4.32).
Proposition (4.32) states that
$$ p \text{ prime } \iff \sigma (p) = p + 1 $$
We need to prove both directions:
($\implies$)
We assume $p$ is prime. That means its only divisors are $p$ and 1. The $\sigma$ function is the sum of divisors, which here is $\sigma(p) = p + 1$.
($\impliedby$)
We start with $\sigma(p)=p+1$. This tells us the divisors of $p$ sum to $p+1$. We know the divisors of $p$ include $p$ and $1$, and these sums to $p+1$. This is the value of $\sigma(p)$ and so there is no possibility for additional divisors. That is, the only divisors of $p$ are $p$ and $1$, and so, by definition, $p$ is prime.
Prove Theorem (4.30).
Theorem (4.30) states that every even perfect number N is of the form:
$$ N = 2^{p−1}(2^p− 1) $$
where $(2^p− 1)$ is prime.
For additional practice, we'll first re-prove Theorem (4.28), which is the converse of Theorem (4.30):
Let $p$ be a prime number. If the Mersenne number $2^p− 1$ is prime then $N = 2^{p−1} (2^p− 1)$ is a perfect number.
We first note that $N$ is a perfect number if $\sigma(N)=2N$, and that $\sigma$ is a multiplicative function.
$$ \begin{align} \sigma(N) &= \sigma \bigl ( \; 2^{p−1} \; \underbrace{(2^p-1)}_{\text{prime}} \; \bigr) \\ \\ & = \sigma(2^{p-1}) \times \sigma(2^p-1) \\ \\ & = (1 + 2 + 2^2 + 2^3 + \ldots + 2^{p-1}) \times (2^p -1 + 1) \\ \\ & = (2^p-1)\times 2^p \\ \\ & = 2 \times 2^{p-1}(2^p-1) \\ \\ &= 2N \end{align} $$
And so we have shown that if $2^p− 1$ is prime then $N = 2^{p−1} (2^p− 1)$ is a perfect number.
We now prove Theorem (4.30).
We start with $N$ as an even perfect number. This means we can write it as $N=2^{k-1}M$, where $k \ge 2$ and $M$ is odd.
Since $N$ is a perfect number, we have $\sigma(N)=2N$. Using that $\sigma$ is multiplicative, we have
$$ \begin{align} 2N & = \sigma(N) \\ \\ 2 \times 2^{k-1}M & = \sigma(2^{k-1}) \times \sigma(M) \\ \\ 2^kM & = (2^k-1) \times \sigma(M) \tag{i} \end{align} $$
This tells us that $2^k \mid (2^k-1) \times \sigma(M)$, but $2^k$ and $(2^k-1)$ are coprime, it must be the case that $2^k \mid \sigma(M)$. That is, for some integer $c$ we have
$$ \sigma(M) = 2^k c $$
Substituting into (i) gives us
$$ \begin{align} 2^kM & = (2^k-1) \times 2^k c \\ \\ M &= (2^k-1) \times c \end{align}$$
This means
$$ N = 2^{k-1} (2^k-1) \times c $$
We need to show $N=2^{p-1}(2^p-1)$, which means we need to show $c=1$.
Let's consider $M = (2^k-1) \times c$ and $\sigma(M) = 2^k c$.
$M$ has divisors $M$ and $c$. We notice that $M+c = (2^k-1) \times c) + c = 2^kc$. This happens to be equal to $\sigma(M)$, telling us that $M$ and $c$ are the only divisors of $M$. Only primes have two divisors, the prime itself and 1. And so $M$ is prime, and $c=1$.
We have shown that every even perfect number $N$ is of the form $2^{p-1}(2^p-1)$, where $(2^p-1)$ is prime.
Note: this solution is inspired by the proof given in Elementary Number Theory (Jones & Jones).
(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
$$ \sigma(p^n) = \frac{p^{n+1}-1}{p-1}$$
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.
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.