Question #260103

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


1
Expert's answer
2021-11-03T10:03:49-0400

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 2p1(2p1)2^{p−1}(2^p − 1) is a perfect number when 2p12^p −1 is prime.

Solution:

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

Divisors of 6=1,2,3,66=1,2,3,6

Divisors of 28=1,2,4,7,14,2828=1,2,4,7,14,28

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

6=1+2+328=1+2+4+7+14\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: 2p12^{p}-1 is prime

To prove: 2p1(2p1)2^{p-1}\left(2^{p}-1\right) is a perfect number

Poof:

Determine all divisors of 2p1(2p1)2^{p-1}\left(2^{p}-1\right) (using that (2p1)\left(2^{p}-1\right) is prime and thus cannot be factorized):

1,2,22,,2p1,(2p1),2(2p1),22(2p1),.2p2(2p1),2p1(2p1)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 2p1(2p1)2^{p-1}\left(2^{p}-1\right) itself.

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

1+2+22++2p1+(2p1)+2(2p1)+22(2p1)++2p2(2p1)=1(1+2p1)+2(1+2p1)+22(1+2p1)++2p2(1+2p1)+2p1=1(2p)+2(2p)+22(2p)++2p2(2p)+2p1=2p(1+2+22++2p2)+2p1\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}

=2p(i=0p22i)+2p1=2p(2p1121)+2p1=2p(2p11)+2p1=2p2p12p+2p1=2p2p122p1+12p1=(2p2+1)2p1=(2p1)2p1=2p1(2p1)\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 2p1(2p1)2^{p-1}\left(2^{p}-1\right) is a perfect number.

Hence Proved


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