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Math Help - Perfect Integers

  1. #1
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    Perfect Integers

    An integer n is called perfect if it equals the sum of all its divisors d 1<d<n
    Let a be a positive integer. Prove that if 2^a -1 is prime, the n= 2^(a-1) (2^a - 1) is perfect.
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    Quote Originally Posted by teramaries View Post
    An integer n is called perfect if it equals the sum of all its divisors d 1<d<n
    Let a be a positive integer. Prove that if 2^a -1 is prime, the n= 2^(a-1) (2^a - 1) is perfect.
    The smallest perfect number is a product of three integers. n obviously has two, but the third is 1, so you can try

    2^{a-1} \cdot (2^a - 1) \cdot 1= 2^{a-1} + (2^a - 1) + 1

    and solve for a. Substitute the number into 2^{a -1} and see if it's prime, and see if n is perfect.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by teramaries View Post
    An integer n is called perfect if it equals the sum of all its divisors d 1<d<n
    Let a be a positive integer. Prove that if 2^a -1 is prime, the n= 2^(a-1) (2^a - 1) is perfect.
    All factors of  n are  \{2^i,\; 2^i\cdot(2^a-1) | 0\leq i<a\}

    So  \sigma(n)=\sum_{d\mid n} d = \sum_{i=0}^{a-1} (2^i+2^i\cdot(2^a-1)) = \sum_{i=0}^{a-1} 2^{i+a} = 2^a\sum_{i=0}^{a-1} 2^i = 2^a\cdot(2^a-1) = 2n , where the last sum is just a geometric sum.
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    Quote Originally Posted by chiph588@ View Post
    All factors of  n are  \{2^i,\; 2^i\cdot(2^a-1) | 0\leq i<a\}
    I have mistaken the vertical bar as a divisor. It took me a while before I realized that it's  \{2^i,\; 2^i\cdot(2^a-1): 0\leq i<a\}
    I did not know what \sigma (n) meant until I read up on the Mersenne prime and Euclid's proof.

    My book has devoted only one chapter on number theory, and I was expected to this proof without much instruction. I am glad you showed the proof.

    I am an engineering student. I learn pure math on my own. Could you recommend me a book on number theory that's easy for my independent study.
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    MHF Contributor chiph588@'s Avatar
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Let p=2^k-1 a prime number.

    Let n=2^{k-1}p.

    gcd(2^{k-1},p)=1

    Now we compute \sigma(n).

    \sigma(n)=\sigma(2^{k-1}p)=\sigma(2^{k-1})\sigma(p)=(2^k-1)(p+1)=(2^k-1)2^k=2n


    An exercise:

    Prove the opposite way.
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  7. #7
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let p=2^k-1 a prime number.

    Let n=2^{k-1}p.

    gcd(2^{k-1},p)=1

    Now we compute \sigma(n).

    \sigma(n)=\sigma(2^{k-1}p)=\sigma(2^{k-1})\sigma(p)=(2^k-1)(p+1)=(2^k-1)2^k=2n
    Beautiful proof.

    An exercise:

    Prove the opposite way.
    That's a good idea. Euler did it.
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    Indeed.
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