Order of 3 modulo a Mersenne prime

**T.Rex** · Mar7-09, 06:45 AM

Hi,

I have the following (new, I think) conjecture about the Mersenne prime numbers, where:

with

prime.
I've checked it up to q = 110503 (M29).

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

With

greatest

such that $LaTeX Code: M_q \\equiv 1 \\pmod{3^i}$ , then we have: $LaTeX Code: O \\leq I$ but no always:

.

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** · Mar14-09, 05:39 PM

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

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

**robert Ihnot** · Mar14-09, 07:22 PM

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:

$LaTeX Code: 4^{22}*4^{248}-1\\equiv 4^{270}-1 \\equiv0 Mod 81$

**T.Rex** · Mar15-09, 03:23 AM

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 paper.
Sorry, the way David found the counter-examples was not so difficult...
Tony

**T.Rex** · Mar15-09, 03:31 AM

Originally Posted by robert Ihnot

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