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Order of 3 modulo a Mersenne prime 
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#1
Mar709, 05:45 AM

P: 62

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 


#2
Mar1409, 04:39 PM

PF Gold
P: 1,059

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. 


#3
Mar1409, 06:22 PM

PF Gold
P: 1,059

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] 


#4
Mar1509, 02:23 AM

P: 62

Order of 3 modulo a Mersenne prime
The conjecture is wrong.
David BroadHurst has found counterexamples. 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 counterexamples was not so difficult... Tony 


#5
Mar1509, 02:31 AM

P: 62

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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