# Proof about perfect numbers and divisors.

1. Dec 20, 2011

### cragar

1. The problem statement, all variables and given/known data
Prove that if $2^{a}-1$ is prime, then $n=2^{a-1}(2^{a}-1)$ is perfect.
3. The attempt at a solution

So by looking at this all divisors of n will be powers of 2 times a prime to the first power or the the zero power. On the $2^{a-1}$ we have (a-1)+1 choices so we have a choices for that divisor and for the prime on the right we have 2 choices. so we have 2a divisors. for the term on the left all the divisors will be $2^0+2^1+2^2 ........2^{a-1}$ so should I do an induction proof to show that this sum equals some formula so that I can show all the positive divisors add up to n?

2. Dec 20, 2011

### Dick

I don't think you need to prove what the sum of the powers of 2 is. It's a geometric series - the formula for summing it is pretty well known.

3. Dec 20, 2011

### cragar

ok, so if n is the product of 2 terms like for example n= $2^x(2^{x+1}-1)$
for the left term i would have x+1 divisors and then for the right term i would have x+2 divisors and then I would need all the divisors of the 2 in combination and then go up to the largest divisor that is not n. is this the right approach.

4. Dec 20, 2011

### Dick

Well, you've got all of the divisors that are powers of two. Can you sum them as a geometric series? Now what are the rest of the divisors? Sum them the same way.

5. Dec 20, 2011

### cragar

thanks for your help by the way.
ok so $2^0+2^1+2^2.......+2^{a-1}=2^a-1$
so summing all the divisors that have powers equals $2^a-1$
now I consider all the the divisors of powers of 2 multiplied by the prime divisor on the right, but I only consider the divisors of all the powers of 2 up to the one right before $2^{a-1}$ because if i considered that one we would have n as a positive divisor and we cant have n be part of the sum for our perfect number.
so we now consider all the divisors of the form $(2^0+2^1+....+2^{a-2})(2^a-1)$
that sum on the left equals $2^{a-1}-1$
so now if we add all the divisors together we get
$(2^a-1)(2^{a-1}-1)+(2^a-1)$
factor out $2^a-1$ and simplify and we get n .

6. Dec 20, 2011

### Dick

That's it alright!

7. Dec 20, 2011