Order of 3 modulo a Mersenne prime

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Discussion Overview

The discussion revolves around a conjecture regarding the order of 3 modulo Mersenne prime numbers, specifically the relationship between the order and the highest power of 3 that divides M_q - 1, where M_q = 2^q - 1 and q is a prime number. The scope includes theoretical exploration and conjectural reasoning based on experimental data.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Tony proposes a conjecture stating that the order of 3 modulo M_q can be expressed as order(3,M_q) = (M_q - 1) / 3^O, where O can be 0, 1, or 2, and relates to the greatest i such that M_q ≡ 1 (mod 3^i).
  • Some participants suggest checking specific Mersenne primes, such as the 27th and 40th, to verify the conjecture's validity through integer results from dividing M_q - 1 by powers of 3.
  • Another participant provides a calculation involving Mersenne prime 27, showing a modular equivalence that supports the conjecture.
  • Tony later retracts the conjecture, stating that David BroadHurst found counter-examples, leading him to propose that the highest power of 3 dividing the order of 3 mod M_q is actually 2.
  • There is a discussion about the implications of the findings, with Tony expressing a desire to clarify his reasoning before making further conjectures.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the conjecture's validity. While some support the initial conjecture with calculations, others point out counter-examples that challenge it, leading to a retraction by Tony.

Contextual Notes

The discussion highlights the complexity of determining the highest power of 3 that divides the order of 3 modulo Mersenne primes, with various assumptions and conditions influencing the claims made. The presence of counter-examples introduces uncertainty regarding the conjecture's accuracy.

T.Rex
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Hi,

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 ?

Tony
 
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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.
 
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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]
 
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" .
Sorry, the way David found the counter-examples was not so difficult...
Tony
 
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robert Ihnot said:
If I understand this correctly we are supposing that 3^3 is the highest dividing power...
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 !).
Thanks,
Tony
 

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