Order of 3 modulo a Mersenne prime

[T.Rex]

Hi,

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

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

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

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

 

[robert Ihnot]

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: \frac{2^{44496}-1}{81} is an integer.

Also, I would suggest trying to check out the 40th Mersenne prime, and find, \frac{2^{20996010}-1}{243} is an integer.

 

[robert Ihnot]

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:

4^{22}*4^{248}-1\equiv 4^{270}-1 \equiv0 Mod 81

 

[T.Rex]

The conjecture is wrong.

The conjecture is wrong.
David BroadHurst has found counter-examples.
The terrible "law of small numbers" has struck again... (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

 

[T.Rex]

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