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
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.
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... (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
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