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Re: Help with Twin Primes

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  • jbrennen
    ... Before somebody corrects me on this, let me rephrase: All EVEN perfect numbers greater than 6 are congruent to 1 (modulo 3). Obviously, an odd perfect
    Message 1 of 4 , Apr 1, 2002
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      --- In primenumbers@y..., "jbrennen" <jack@b...> wrote:

      > All perfect numbers greater than 6 are congruent to 1 (modulo 3)

      Before somebody corrects me on this, let me rephrase:

      All EVEN perfect numbers greater than 6 are congruent
      to 1 (modulo 3).

      Obviously, an odd perfect number (if such exists) cannot be the
      number in the middle of a twin prime pair.

      As a side note, there are probably no odd perfect numbers, so
      the original statement was probably true, but until somebody
      proves the nonexistence of odd perfect numbers, I must watch
      myself :-)
    • Steven Whitaker
      ... There aren t any others. [Apologies to the real mathematicians for the standard of the proof. I hope the logic is clear] Given that these 3 numbers must be
      Message 2 of 4 , Apr 1, 2002
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        On Mon, 01 Apr 2002 21:38:27 -0000, you wrote:

        >I am looking for all twin primes such that their arithmetic mean is a
        >perfect number.
        >
        >For example:
        >
        >If 5 and 7 are our primes then 6 is our arithmetic mean.
        >
        >Thank you for your help.


        There aren't any others.

        [Apologies to the real mathematicians for the standard of the proof. I
        hope the logic is clear]

        Given that these 3 numbers must be adjacent integers, then one of them
        must be divisible by 3. Neither of the primes is 3 (because no number
        adjacent to this is perfect), so it must be the perfect number that is
        divisible by 3. It must be an even perfect number (because otherwise the
        primes would both be even) and is therefore the product of a power of
        two and a Mersenne prime. Obviously a power of two is not divisible by
        three and so the Mersenne prime must be 3. And the only perfect number
        of this form is 6, which gives you the solution you already have.

        Regards
        Steve nice but dim
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