Question

Prove the following:
If p is prime and n is any positive integer, then sum of divisors of (p^n) is (p^(n+1)-1)/(p-1). Prove the following:
If p is prime and n is any positive integer, then sum of divisors of (p^n) is (p^(n+1)-1)/(p-1). @Mathematics

Answer 1

Are we including 1 and p^n?

Answer 2

You have to do a proof by induction for this one. First, we need to state that the sum of all the divisors of \[p^n\] are \[1+ p+ p^2+ p^3+ ...+ p^n\] since p is prime. Then we start the proof by induction. First, let \[n = 0\] Then \[p^0 = 1 = {p^{0+1} - 1 \over p - 1}\] Then we assume this is true for some \[k \in \mathbb{N}\] so \[1+ p + p^2+...+p^k = {p^{k+1} - 1 \over p - 1}\] Then\[1+p+...p^k+p^{k+1} = {p^{k+1} - 1 \over p - 1} + p^{k+1} = {(p^{k+1} - 1) + p^{k+1}(p - 1) \over p - 1}\] So \[1+p+...+p^{k+1} = {p^{(k+1)+1} + p^{k+1}-p^{k+1} -1 \over p -1} = {p^{(k+1)+1} - 1 \over p-1}\] Which is what we wanted so therefore, the sum of the divisors of \[p^n\] where p is prime, is \[{p^{n+1} - 1 \over p -1} \forall n \in \mathbb{N}\]

Answer 3

trying to use this one to prove the other one I asked before. I can't justify how the sum of the divisors 2^(k-1) is (2^(k+1)-1)

Answer 4

Where do you need the sum of the divisors of 2^(k-1)?

Answer 5

http://primes.utm.edu/notes/proofs/EvenPerfect.html in this proof

Answer 6

Wouldn't you be trying to show the sum of the divisors of 2^(k-1) is ((2^k) -1)?

Answer 7

I can easily say that the sum of the divisors of a prime (2^k)-1 is just 2^k since it's just 1+ (2^k)-1

but the sum of the divisors of 2^(k-1) is a lot harder to show

Answer 8

yes you are right sorry

Answer 9

Well, shouldn't that be almost trivial now since 2 is prime? The factors of 2^(k-1) should sum to \[{2^k -1 \over 2-1} = 2^k -1\] by what we just proved.

Answer 10

the problem with this way is that it's assuming that I know what the form of the sum should be in and almost proving it backwards. It's like I'm assuming something I shouldn't know. Is there a simpler way to show that 2^(k-1) is ((2^k) -1)?

Answer 11

maybe a way to show that it equals this in this specific case

Answer 12

To the best of my knowledge, that's the best way to do it.

Answer 13

We could do another proof by induction using 2 instead of p, but that's just a degenerate case of what we proved earlier.

Answer 14

the thing is that I'm supposed to build the proofs off of theorems and propositions I already know. So I can't just go and prove it by knowing the conclusion. The reason why I asked this question was so I could understand where it was coming from better so I could maybe derive it myself without first knowing the form of the conclusion

Answer 15

You could try making some sort of combinatorial argument, but I'm not sure how well it work. I can't help you much more.

Answer 16

can you prove this by induction?
If k is a positive integer and (2^k)-1 is prime, then (2^(k-1))*((2^k)-1) is perfect.

Answer 17

I was think that it might be possible to do it that way since the conclusion is would be a sum of numbers which is a form that induction would be good for.

Answer 18

I don't think that can be proven by induction because 2^(k+1) - 1 is not necessarily prime.

Answer 19

Good luck!

Answer 20

thanks for your help, I will do my best!