# Thread: proof with perfect and prime numbers

1. ## proof with perfect and prime numbers

problem: given that the divisors of 2^(n-1)p are 1,2,2^2,...,2^(n-1) and p,2p,(2^2)p,...,2^(n-2)p; show that 2^(n-1)p is perfect when p=2^(n) -1 is prime.

to show that 2^(n-1) p is perfect we need to show that it is equal to the sum of its divisors (including 1 but excluding itself).
so we would show that
2^(n-1)p = 1+2+2^2+...+2^(n-1) + p+2p+2^(2)p+...+2^(n-2)p
2^(n-1)p = 1+2+2^2+...+2^(n-1) + p [1+2+2^(2)+...+2^(n-2)]

* 1+2+2^2+...+2^(n-1) = 2^(n) -1 * therefore:

2^(n-1)p = 2^(n)-1 + p [1+2+2^(2)+...+2^(n-2)]

i dont know how to manipulate this equation anymore to get the right side to look like the left side?

any suggestions?

2. Originally Posted by jean91890
problem: given that the divisors of 2^(n-1)p are 1,2,2^2,...,2^(n-1) and p,2p,(2^2)p,...,2^(n-2)p; show that 2^(n-1)p is perfect when p=2^(n) -1 is prime.

to show that 2^(n-1) p is perfect we need to show that it is equal to the sum of its divisors (including 1 but excluding itself).
so we would show that
2^(n-1)p = 1+2+2^2+...+2^(n-1) + p+2p+2^(2)p+...+2^(n-2)p
2^(n-1)p = 1+2+2^2+...+2^(n-1) + p [1+2+2^(2)+...+2^(n-2)]

* 1+2+2^2+...+2^(n-1) = 2^(n) -1 * therefore:

2^(n-1)p = 2^(n)-1 + p [1+2+2^(2)+...+2^(n-2)]

i dont know how to manipulate this equation anymore to get the right side to look like the left side?

any suggestions?
You need to remember the formulae for the sum of geometric sequences, and take into account that the divisors of

$\displaystyle 2^{n-1}p=2^{n-1}(2^n-1)=2^{2n-1}-2^{n-1}$ are the ones given to you above, so their sum is:

$\displaystyle \sum\limits^{n-1}_{k=0}2^k+p\sum\limits^{n-2}_{k=0}2^k=2^n-1+p(2^{n-1}-1)=p+p(2^{n-1}-1)=2^{n-1}p$

Tonio