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Order of 3 modulo a Mersenne prime

  1. Mar 7, 2009 #1

    I have the following (new, I think) conjecture about the Mersenne prime numbers, where: [tex]M_q = 2^q - 1[/tex] with [tex]q[/tex] prime.
    I've checked it up to q = 110503 (M29).

    Conjecture (Reix): [tex]\large \ order(3,M_q) = \frac {M_q - 1}{3^O}[/tex] where: [tex]\ \large O = 0,1,2[/tex] .

    With [tex]I =[/tex] greatest [tex]i[/tex] such that [tex]M_q \equiv 1 \pmod{3^i}[/tex] , then we have: [tex]O \leq I[/tex] but no always: [tex]O = I[/tex] .

    A longer description with experimental data is available at: ConjectureOrder3Mersenne.

    Samuel Wagstaff was not aware of this conjecture and has no idea (yet) about how to prove it.

    I need a proof...
    Any idea ?

  2. jcsd
  3. Mar 14, 2009 #2
    If I understand this correctly we are supposing that 3^3 is the highest dividing power, but take the 27th Mersenne prime, as shown in a table, and consider: [tex]\frac{2^{44496}-1}{81} [/tex] is an integer.

    Also, I would suggest trying to check out the 40th Mersenne prime, and find, [tex]\frac{2^{20996010}-1}{243}[/tex] is an integer.
    Last edited: Mar 14, 2009
  4. Mar 14, 2009 #3
    T.Rex: I've checked it up to q = 110503 (M29)

    If you want to see some check work on Mersenne 27, notice that 2^2000==4 Mod 81.

    Thus dividing out 44496/2000 = 22 + Remainder 496. 496 = 2*248. Thus:

    [tex]4^{22}*4^{248}-1\equiv 4^{270}-1 \equiv0 Mod 81 [/tex]
  5. Mar 15, 2009 #4
    The conjecture is wrong.

    The conjecture is wrong.
    David BroadHurst has found counter-examples.
    The terrible "law of small numbers" has struck again... :cry::mad::confused::frown: (but the numbers were not so small...).
    I've updated the paper and just conjectured that the highest power of 3 that divides the order of 3 mod M_q is 2. But it is not so much interesting...
    Never mind, we learn by knowing what's false too.
    I've updated the http://tony.reix.free.fr/Mersenne/ConjectureOrder3Mersenne.pdf" [Broken].
    Sorry, the way David found the counter-examples was not so difficult...
    Last edited by a moderator: May 4, 2017
  6. Mar 15, 2009 #5
    Not exactly, Robert. For q=44497, 4 is the highest power of 3 that divides Mq-1, but 1 is the highest power of 3 in the relationship between (Mq-1) and order(3,Mq).
    I have other reasons to think that 2 is the highest power of 3 in this relationship. But I need to clarify that before conjecturing again (one mistake is enough !!).
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