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?