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Proof about perfect numbers and divisors.

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that if [itex] 2^{a}-1 [/itex] is prime, then [itex] n=2^{a-1}(2^{a}-1) [/itex] 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 [itex] 2^{a-1} [/itex] 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 [itex] 2^0+2^1+2^2 ........2^{a-1} [/itex] 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?
     
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  3. Dec 20, 2011 #2

    Dick

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    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.
     
  4. Dec 20, 2011 #3
    ok, so if n is the product of 2 terms like for example n= [itex] 2^x(2^{x+1}-1) [/itex]
    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.
     
  5. Dec 20, 2011 #4

    Dick

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    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.
     
  6. Dec 20, 2011 #5
    thanks for your help by the way.
    ok so [itex] 2^0+2^1+2^2.......+2^{a-1}=2^a-1 [/itex]
    so summing all the divisors that have powers equals [itex] 2^a-1 [/itex]
    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 [itex] 2^{a-1} [/itex] 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 [itex] (2^0+2^1+....+2^{a-2})(2^a-1) [/itex]
    that sum on the left equals [itex] 2^{a-1}-1 [/itex]
    so now if we add all the divisors together we get
    [itex] (2^a-1)(2^{a-1}-1)+(2^a-1) [/itex]
    factor out [itex] 2^a-1 [/itex] and simplify and we get n .
     
  7. Dec 20, 2011 #6

    Dick

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    That's it alright!
     
  8. Dec 20, 2011 #7
    sweet thanks for your help
     
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