Proof about perfect numbers and divisors.

[cragar]

Homework Statement

Prove that if $2^{a}-1$ is prime, then $n=2^{a-1}(2^{a}-1)$ is perfect.

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?

 

[Dick]

Homework Statement

Prove that if $2^{a}-1$ is prime, then $n=2^{a-1}(2^{a}-1)$ is perfect.

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?

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.

 

[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.

 

[Dick]

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.

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.

 

[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 can't 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 .

 

[Dick]

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 can't 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 .

That's it alright!

 

[cragar]

sweet thanks for your help