Example Questions Regarding k-Perfect Ab. and Def. Numbers

# Example Questions Regarding k-Perfect Abundant and Deficient Numbers

Before we look at the following questions, let's first recall the following definitions for a number n such that $n \in \mathbb{Z^+}$:

• 1) n is considered perfect if $\sigma (n) = 2n$.
• 2) n is considered k-perfect if $\sigma (n) = kn$.
• 3) n is considered abundant if $\sigma (n) > 2n$.
• 4) n is considered deficient if $\sigma (n) < 2n$.

We will now look at some questions regarding k-perfect, abundant, and deficient numbers.

## Example 1

Suppose that $n \in \mathbf{Z^+}$ and is odd and is also 4-perfect. Determine what value of k makes 4n k-perfect.

We first note that since n is odd, 2 does not appear in the prime power decomposition of of n. Suppose that the prime power decomposition of $n = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$. It thus follows that the prime power decomposition of $4n = 2^2 p_1^{e_1}p_2^{e_2}...p_k^{e_k}$. Hence:

(1)
\begin{align} \sigma(4n) = \sigma(2^2)\sigma (p_1^{e_1}p_2^{e_2}...p_k^{e_k}) \\ \sigma(4n) = 7\sigma (n) \\ \sigma (4n) = 7(4n) \\ \end{align}

Hence if n is an odd positive integer and is 4-perfect, then 4n is a 7-perfect number. Hence k = 7.

## Example 2

Prove that if n is odd, then $\phi(2n) = \phi (n)$.

Suppose that n is odd. It thus follows that $(2, n) = 1$, and hence $\phi (2n) = \phi (2) \phi (n) = \phi (n)$ (since $\phi (2) = 1$). Hence the proof is complete.

## Example 3

Prove that if n is even, then $\phi (2n) = 2 \phi (n)$.

If n is even, then n can be written such that $n = 2^em$ where e ≥ 1. Hence it follows that since $(2, m) = 1$, then $\phi (n) = \phi (2^e) \phi (m) = 2^{e - 1} \phi (m)$.

Now we will look at 2n. We notice that $2n = 2^{e+1}m$, and since $(2, m) = 1$, it follows that $\phi (2n) = 2^e \phi (m)$. Taking out a factor of 2, we get that $\phi (2n) = 2 \cdot 2^{e - 1} \phi (m)$, and we can make the following substitution that $\phi (2n) = 2 \phi (n)$.

## Example 4

If n is a 3-perfect number and $(n, 12) = 1$, then determine if 12n is a k-perfect number.

We note that since $(n, 12) = 1$, it follows that $\phi (12n) = \phi(12) \phi (n) = \phi(4) \phi (3) \phi (n) = 7 \cdot 4 \phi (n)$.

But we note that since n is a 3-perfect number, then $\phi (n) = 3n$. Hence it follows that $\phi (12n) = 7 \cdot 4 \cdot 3n = 7 \cdot 12n$. Hence 12n is a 7-perfect number.

## Example 5

If n is a 7-perfect number and $(n, 14) = 1$, then determine if 14n is a k-perfect number.

Since $(n, 14) = 1$, it follows that $\phi (14n) = \phi (14) \phi (n) = \phi (2) \phi (7) \phi (n) = 3 \cdot 8 \cdot \phi (n)$.

Since n is a 7-perfect number, then $\phi (n) = 7n$. Hence it follows that $\phi (14n) = 3 \cdot 8 \cdot 7n = 3 \cdot 4 \cdot 14n = 12 \cdot 14n$. Hence 14n is a 12-perfect number.

## Example 6

If p is prime, n is a p-perfect number, and $(n, p) = 1$, then determine if pn is a k-perfect number.

Since $(n, p) = 1$, it follows that $\phi (pn) = \phi (p) \phi (n) = (p + 1) \phi (n)$.

Since n is a p-perfect number, it follows that $\phi(n) = pn$. Hence $\phi (pn) = (p + 1)pn$. So pn is a (p + 1)-perfect number.

## Example 7

For what values of n is $m = 2^n \cdot 11$ classified as abundant?

We first note that $(2^n, 11) = 1$, hence it follows that $\sigma (m) = \sigma (2^n \cdot 11) = \sigma (2^n)\sigma(11) = (2^{n+1} - 1) \cdot 12$.

Furthermore, we know that $2m = 2 \cdot 2^n \cdot 11 = 2^{n + 1} \cdot 11$.

Hence we want to solve the inequality $12\cdot 2^{n+1} - 12 > 11 \cdot 2^{n + 1}$. Subtracting $11 2^{n+1}$ from both sides, we get $2^{n + 1} - 12 > 0$, or rather $2^{n + 1} > 12$. This only happens when n > 3.