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    Perfect number

    Prove that, $\displaystyle {2^{(p - 1)}}({2^p} - 1)$ is perfect number if $\displaystyle ({2^p} - 1)$ 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, $\displaystyle {2^{(p - 1)}}({2^p} - 1)$ is perfect number if $\displaystyle ({2^p} - 1)$ is prime number
    write the definition of the a perfect number.
    now answer this question:
    list the factors of the number $\displaystyle 2^kp$, where $\displaystyle p$ 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, $\displaystyle {2^{(p - 1)}}({2^p} - 1)$ is perfect number if $\displaystyle ({2^p} - 1)$ is prime number
    A number is perfect if the sum of it's divisors are twice the number.

    Since by assumption

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

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

    $\displaystyle 1,2,4,...2^{p-1}$

    Since the latter is a geometric series we get that

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

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

    $\displaystyle (2^p-1)+(2^p-1)=2(2^p-1)$
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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, $\displaystyle {2^{(p - 1)}}({2^p} - 1)$ is perfect number if $\displaystyle ({2^p} - 1)$ is prime number
    Also, you should know that if $\displaystyle n$ is an even perfect number say $\displaystyle n=2^{k-1}m$ then $\displaystyle m=2^k-1$ and it is a prime number.

    Proof:

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

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

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

    $\displaystyle n$ is perfect number, therefor:

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

    With (1), we get:

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

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

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

    Or:

    $\displaystyle 2^kM=\sigma(m)$.

    $\displaystyle m$ and $\displaystyle M$ are both divisors of m, and therefor:

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

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

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

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