Answer to Question #260103 in Combinatorics | Number Theory for King

Question

We call a positive integer perfect if it equals the sum of its positive divisors

other than itself.

(a) Prove that 6 and 28 are perfect numbers

(b) Prove that if 2p − 1 is prime, then 2p−1

(2p − 1) is a perfect number

Answer 1

Question is not written in proper way.

Actual question is:

We call a positive integer perfect if it equals the sum of its positive divisors other than itself.

a) Prove that 6 and 28 are perfect.

b) Prove that $2^{p−1}(2^p − 1)$ is a perfect number when $2^p −1$ is prime.

Solution:

(a) Let us determine all divisors of each number.

Divisors of $6=1,2,3,6$

Divisors of $28=1,2,4,7,14,28$

Let us determine the sum of all positive divisors other than the integer itself:

$\begin{gathered} 6=1+2+3 \\ 28=1+2+4+7+14 \end{gathered}$

We then note that 6 and 28 are both perfect.

(b) Given: $2^{p}-1$ is prime

To prove: $2^{p-1}\left(2^{p}-1\right)$ is a perfect number

Poof:

Determine all divisors of $2^{p-1}\left(2^{p}-1\right)$ (using that $\left(2^{p}-1\right)$ is prime and thus cannot be factorized):

$1,2,2^{2}, \ldots, 2^{p-1},\left(2^{p}-1\right), 2\left(2^{p}-1\right), 2^{2}\left(2^{p}-1\right), \ldots .2^{p-2}\left(2^{p}-1\right), 2^{p-1}\left(2^{p}-1\right)$

Note: The last divisor is the number $2^{p-1}\left(2^{p}-1\right)$ itself.

Let us determine the sum of all positive divisors other than the integer itself:

$\begin{aligned} &1+2+2^{2}+\ldots+2^{p-1}+\left(2^{p}-1\right)+2\left(2^{p}-1\right)+2^{2}\left(2^{p}-1\right)+\ldots+2^{p-2}\left(2^{p}-1\right) \\ &=1\left(1+2^{p}-1\right)+2\left(1+2^{p}-1\right)+2^{2}\left(1+2^{p}-1\right)+\ldots+2^{p-2}\left(1+2^{p}-1\right)+2^{p-1} \\ &=1\left(2^{p}\right)+2\left(2^{p}\right)+2^{2}\left(2^{p}\right)+\ldots+2^{p-2}\left(2^{p}\right)+2^{p-1} \\ &=2^{p}\left(1+2+2^{2}+\ldots+2^{p-2}\right)+2^{p-1} \end{aligned}$

$\begin{aligned} &=2^{p}\left(\sum_{i=0}^{p-2} 2^{i}\right)+2^{p-1} \\ &=2^{p}\left(\frac{2^{p-1}-1}{2-1}\right)+2^{p-1} \\ &=2^{p}\left(2^{p-1}-1\right)+2^{p-1} \\ &=2^{p} \cdot 2^{p-1}-2^{p}+2^{p-1} \\ &=2^{p} \cdot 2^{p-1}-2 \cdot 2^{p-1}+1 \cdot 2^{p-1} \\ &=\left(2^{p}-2+1\right) 2^{p-1} \\ &=\left(2^{p}-1\right) 2^{p-1} \\ &=2^{p-1}\left(2^{p}-1\right) \end{aligned}$

By the definition of a perfect number, we have then shown that $2^{p-1}\left(2^{p}-1\right)$ is a perfect number.

Hence Proved