Mathematics / Mathematik / Matemática

Mathematics / Mathematik / Matemática

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  • Winfried Weber
    Winfried Weber    Group moderator
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    Mersenne numbers
    Hallo,
    I have some questions to the Mersenne Numbers M(n) := 2^n-1, (n€N)

    The Mersenne numbers are “important” because the greatest known prime numbers are Mersenne Numbers.

    I would like to start with:

    The following facts are well known:
    - M(n) is a prime number, only if n is a prime.
    -The controverse is wrong : M(11) = 2^11 – 1 = 2047 = 23 * 89
    - if n=r*s , then M(r) and M(s) are divisors of M(n).

    Regard now:
    let p a prime number, p>=3, then q:= (p-1)/2 is a natural number.
    So M(2)=3 and M(q) are divisors of M(p-1).

    the Conjecture:
    “p is a prime divisor of M(p-1).”
    The conjecture is true for p=3, 5, 7, 11, … , 47. (I did not yet try further numbers.)

    Problem:
    If we factorize M(p-1) = M(q) * z with a natural number z,
    then “sometimes” p is a prime divisor of M(q), sometimes p is a divisor of z. (at least for 3=<p=<47)

    Examples:
    p=17, q=8
    M(16) = 3*5*17*257
    M(8)= 3*5*17, so p=17 prime divisor of M(q)

    p=19, q=9
    M(18) = 3*3*3*19*511
    M(9)= 511, so p=19 is prime divisor of z=3*3*3*19


    If the conjecture is true:
    What is the condition for p to be a prime divisor of M(q)?

    If the conjecture is not true:
    (smallest) counterexample?


    Best regards,
    Winfried
    • 19 Apr 2007, 11:37 am
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  • Winfried Weber
    Winfried Weber    Group moderator
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    Re^2: Mersenne numbers
    Christian Niemetz schrieb:
    Hi Winfried,
     
    Regarding your conjecture, that “p is a prime divisor of M(p-1)” I can tell you that this is true for all p with p prime.
     
    It is a direct conclusion from the so-called "little FERMAT":
          Hello Christian,
    you are right, that was quite easy (Why I didn't see that?).

    We have for p€ P:={ p€ N, p prime, p>=3} , q:= (p-1)/2

    M(p-1) = (2^q-1) * (2^q+1) = M(q) * [M(q)+2]

    Define
    P(1) : = { p€ P, p is divisor of M(q)}
    P(2) : = { p€ P, p is divisor of M(q)+2}

    P(1)={7, 17, 23, 31, 41, ...}
    P(2)={3, 5 , 11, 13, 19, 29, 37, ...}

    P(1) and P(2) have no commen elements, P(1) and P(2) are together P ("little Fermat").


    QUESTION:
    Is it possible to discribe P(1) and P(2) in an easier way?

    CONJECTURE:
    p= 1 mod 8 or p = 7 mod 8 => p € P(1)
    p= 3 mod 8 or p = 5 mod 8 => p € P(2)


    LG,
    Winfried
    • 20 Apr 2007, 10:39 am
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  • Dr. Andreas Ecker
    Dr. Andreas Ecker
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    Re^4: Mersenne numbers
    Hi,

    I agree with Christian.

    My argumentation would be as follows:
    Using the law of quadratic reciprocity (see http://en.wikipedia.org/wiki/Quadratic_reciprocity or http://de.wikipedia.org/wiki/Quadratisches_Reziprozit%C3%A4t...) we know that 2 is a square in F_p (the field with p elements), iff p is congruent to 1 or 7 modulo 8.
    Every element in F_p - {0} is a solution of the equation
    x^(p-1) - 1 = 0
    If 2 is a square in F_p, then there is y in F_p, so that y^2=2.
    Then 2^q = y^(2q) = y^(p-1) = 1.
    On the other hand, if 2^q=1, then 2 has to be a square, as the multiplicative group of a finite field is cyclic.

    Regards,
    Andreas
    • 23 Apr 2007, 8:20 pm