**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**

The **Answer to the Question**

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**Here's the Solution to this Question**

a.

The divisors of $6$ are $1,2, \space and \space 3$. If we sum these divisors then we have ,

$1+2+3=6$, Therefore, 6 is a perfect number.

The divisors of $28$ are $1,2,4,7,14$. Summing these divisors gives,

$1+2+4+7+14=28$. This shows that $28$ is a perfect number.

b.

We need to show that If $2^p-1$ is a prime number, then $2^{p-1}(2^p-1)$ is a perfect number.

Let $p$ be an integer so that $2^p-1$ is a prime number. Also, let $n=2^p-1$ so that $2^{p-1}(2^p-1)=2^{p-1}n$. We can write the divisors of $2^{p-1}$ as, $1,2,4,8,...,2^{p-1}$. Since $n$

is prime, we can write out the other divisors of $2^{p-1}n$ as, $n,2n,4n,8n,...,2^{p-2}n$.

Let us recall that,

$1+x+x^2+x^3+...+x^{p-1}=(x^p-1)/(x-1)$, where $x=2$ so,

$1+2+4+8+...+2^{p-1}=(2^p-1)/(2-1)=2^p-1=n...........(i)$

We can use the same formula we have used to find $equation(i)$ for,

$n+2n+4n+8n+...+2^{p-2}n$

We can see that,

$n+2n+4n+8n+...+2^{p-2}n=n(1+2+4+8+...+2^{p-2})=n(1+2^1+2^2+2^3+...+2^{p-2})=n(2^{p-1}-1)/(2-1)=n(2^{p-1}-1)............(ii)$

We can now sum all the divisors of $2^{p-1}n$ by combining equations $(i)$ and $(ii)$, $1+2+4+...+2^{n-1}+n+2n+4n+...+2^{p-2}n=n+n(2^{p-1}-1)=n+2^{p-1}n-n=2^{p-1}n=2^{p-1}(2^p-1)$

Therefore, if $2^p-1$ is prime then $2^{p-1}(2^p-1)$ is perfect. $\blacksquare$