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    Perfect number

    Prove that, {2^{(p - 1)}}({2^p} - 1) is perfect number if ({2^p} - 1) is prime number
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by seit View Post
    Prove that, {2^{(p - 1)}}({2^p} - 1) is perfect number if ({2^p} - 1) is prime number
    write the definition of the a perfect number.
    now answer this question:
    list the factors of the number 2^kp, where p is prime.
    then sum of the factors.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by seit View Post
    Prove that, {2^{(p - 1)}}({2^p} - 1) is perfect number if ({2^p} - 1) is prime number
    A number is perfect if the sum of it's divisors are twice the number.

    Since by assumption

    2^p-1 is prime the has only two divisors 1 and itself.

    Since the other factor is just powers of 2 its divisors are

    1,2,4,...2^{p-1}

    Since the latter is a geometric series we get that

    \sum_{n=1}^{p}2^{n-1}=2^p-1

    Since the has already added the divisor 1 we just add other divisor

    (2^p-1)+(2^p-1)=2(2^p-1)
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by seit View Post
    Prove that, {2^{(p - 1)}}({2^p} - 1) is perfect number if ({2^p} - 1) is prime number
    Also, you should know that if n is an even perfect number say n=2^{k-1}m then m=2^k-1 and it is a prime number.

    Proof:

    m\in \mathbb{N} odd, k\geqslant 2 .

    \text{gcd}(2^{k-1},m)=1, hence:

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

    n is perfect number, therefor:

    \sigma(n)=2n=2^km

    With (1), we get:

    2^km=(2^k-1)\sigma(m) (2)

    From the above we can deduce that m=(2^k-1)M, now if we put this m to (2) we will get:

    2^k(2^k-1)M=(2^k-1)\sigma(m)

    Or:

    2^kM=\sigma(m).

    m and M are both divisors of m, and therefor:

    2^kM=\sigma(m)\geqslant m+M=2^kM.

    The conclusion is that: \sigma(m)=m+M.

    From the above follows that m is prime number and M=1

    Or: m=(2^k-1)M=2^k-1
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