# Order of 3 modulo a Mersenne prime

by T.Rex
Tags: mersenne, modulo, order, prime
 P: 62 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
 PF Patron 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: $$\frac{2^{44496}-1}{81}$$ 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.
 PF Patron 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: $$4^{22}*4^{248}-1\equiv 4^{270}-1 \equiv0 Mod 81$$
P: 62

## Order of 3 modulo a Mersenne prime

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
P: 62
 Quote 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

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