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Perfect numbers beside 6 in mod6.

  1. May 6, 2006 #1
    i chekced a few perfect numbers with module 6 and, a nice property is that all mod 6 equal 4 (at least for those i checked), i guees that if an odd perfect number would exist then its mod 6 would be different.

    i wonder how to prove that for every even perfect number greater than 6, its mod 6 equal 4?

    i guess because its even it's divisble by 2, and then the question becomes how to prove that mod 3 equal 2, then how do you prove/disprove the assertion?
     
  2. jcsd
  3. May 6, 2006 #2

    matt grime

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    obviously: any number congruent to 0,2,4 mod 6 is even.

    it seems an interesting conjecture.
     
  4. May 6, 2006 #3
    All even perfect numbers are of the form 2^(n - 1) * (2^n - 1) where n is prime, so if we restrict our attention to the case where n > 2, we see that

    2^(n - 1) * (2^n - 1) = (-1)^(n - 1) * ((-1)^n - 1) = 1(-1 -1) = 1 (mod 3)

    since n - 1 is even and n is odd.
     
  5. May 6, 2006 #4

    shmoe

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    Small correction bolded.

    http://primes.utm.edu/notes/proofs/EvenPerfect.html for a proof of Muzza's statement about even perfect numbers if you haven't seen one yet (in most elementary texts as well).
     
  6. May 6, 2006 #5
    muzza, why this "2^(n - 1) * (2^n - 1) = (-1)^(n - 1) * ((-1)^n - 1)"?
    or you were reffering to conguerence here, and even if you did refer to conguernece shouldn't it be mod2
    because:
    2^(2n-1)-2^(n-1)-(-1)^(2n-1)-(-1)^n=2^(2n-1)-2^(n-1)+(-1)^2n+(-1)^(n+1)=2^(2n-1)-2^(n-1)+2=0mod2
     
  7. May 6, 2006 #6
    All the calculations were made modulo 3, as indicated by the "(mod 3)" at the end of the line.

    Why would you be interested in working modulo 2 when you want to prove that something is equal to something else modulo 3...?
     
  8. May 7, 2006 #7
    then how do you infer that: 2^(2n-1)-2^(n-1)+2 is mod3.
     
  9. May 7, 2006 #8

    Gokul43201

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    [tex]2 \equiv -1~~ (mod~3)~~\implies~2^a \equiv -1^a~~(mod~3) [/tex]
     
  10. May 7, 2006 #9

    Gokul43201

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    There must be a much nicer way to do this ...

    First I assert that [itex]4^n \equiv 4~(mod~6) [/itex] ---(1)

    Proof (1): (whenever I've missed it, everything that follows is a congruence mod 6)

    [tex]4^n \equiv (-2)^n = (-1)^n \cdot 2^n [/tex]

    Since 2^n is not divisible by 6, 2^n = 6m+2 or 6m+4. Next, notice that no two consecutive powers of 2 can take the same form, for then we'd have :

    [tex]2^n = 2^{n+1} - 2^n = (6m'+2) - (6m+2) = 6q~[/tex], which is not possible (and likewise with the form 6m+4).

    This shows that 2^n must alternate between the above two forms. In other words, modulo 6, 2^n must alternate between 4 and -4. Since [itex]2^1 \equiv -4~(mod~6) [/itex], we have [itex]2^n \equiv (-1)^n\cdot 4~(mod~6) [/itex] and hence [itex](-1)^n\cdot 2^n \equiv 4 ~(mod~6) [/itex]

    This proves assertion 1.

    Next we simply note that an even perfect number can be written as

    [tex]P = 2^{p-1} \cdot(2^p-1) [/tex]

    For odd p > 2, we write p=2n+1, which gives

    [tex]P = 2^{2n} \cdot (2^{2n+1} -1 ) = 4^n (2\cdot 4^n -1) \equiv 4(4\cdot2-1) \equiv 4 [/tex]
     
    Last edited: May 7, 2006
  11. May 7, 2006 #10
    Huh? 2^(2n-1)-2^(n-1)+2 is a number, one can perform calculations with it modulo 3 without needing to know that the number itself "is mod 3" (which makes no sense anyway...).
     
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