 mersenneforum.org > Math How often is 2^p-1 square-free
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read  2005-12-07, 22:45 #1 Zeta-Flux   May 2003 7×13×17 Posts How often is 2^p-1 square-free If p is prime, how often is 2^p-1 square-free? Here is the background to my question: Fix a prime q. It is well known that a^{q-1} is congruent to 1 mod q, if q does not divide a. However, it can also happen that a^{q-1} is congruent to 1 mod q^2. We can in fact find all such a (up to a multiple of q^2) rather easily. We can continue this process, with q^3, q^4, etc... The size of possible a's grows at about the same rate. So, if we fix a equal to some number (like 2) it seems unlikely that 2^p will be congruent to 1 mod q^2 (where p | (q-1)) for large q. Is this a known result, or is my heuristic flawed?   2005-12-07, 23:41   #2
John Renze

Nov 2005

24×3 Posts Quote:
 Originally Posted by Zeta-Flux If p is prime, how often is 2^p-1 square-free?
It is conjectured that all Mersenne numbers are squarefree.

http://primes.utm.edu/notes/proofs/SquareMerDiv.html

John   2005-12-07, 23:46   #3
Dougy

Aug 2004
Melbourne, Australia

23×19 Posts Quote:
 If p is prime, how often is 2^p-1 square-free?
This is currently an open question. In fact every Mersenne number with prime exponent examined so far has been square free. The following is known.

Definition: A prime such that is known as a Wieferich prime.

Theorem: Let be a prime such that divides for some prime . Then and so is a Wieferich prime.

Proof: Since divides , we have for some natural number . So . Therefore .

The only known Wieferich primes are 1093 and 3511 and there are no others less than . Crandall, Dilcher, Pomerance, Math. Comp., 1997.

Something I'd also like to point out here is that if we didn't restrict the exponent to primes then there are lots of non-square-free Mersenne numbers. Let be an odd natural number, then divides for all natural numbers . This follows straight from the Euler-Fermat theorem.

While it has been conjectured that all Mersenne numbers with prime exponents are square-free, it has also been conjectured that not all are square-free. Guy, Unsolved Problems in Number Theory, Springer, 1994. For this reason I describe it as an "open question," rather than a conjecture.   2005-12-08, 04:01 #4 Zeta-Flux   May 2003 110000010112 Posts Great. That is just the type of thing I'm looking for. What I'd really like to know is if there is a similar finite set of examples where q^2 | 3^p-1, and a large lower bound they have tested up to. (Same thing for 7^p -1 and 11^p -1...etc...)   2005-12-12, 04:06   #5
wblipp

"William"
May 2003
New Haven

93916 Posts Quote:
 Originally Posted by Zeta-Flux What I'd really like to know is if there is a similar finite set of examples where q^2 | 3^p-1, and a large lower bound they have tested up to. (Same thing for 7^p -1 and 11^p -1...etc...)
I know of three examples of squares dividing p^q-1 for odd p and q less than 100

11^2 | 3^5
7^2 | 67^3
47^2 | 71^23

I also have some examples for larger values of p if that's of any interest. Also a few examples of cubed divisors.

Last fiddled with by wblipp on 2005-12-12 at 04:09   2005-12-12, 04:31 #6 Citrix   Jun 2003 30538 Posts You forgot 3^2-1 %(2^3)==0 Only solution to a^x-b^y=1 Citrix   2005-12-12, 11:00 #7 Zeta-Flux   May 2003 7·13·17 Posts William, Thanks. I found a paper by Peter Montgomery, giving the bounds I needed. Cheers, Pace   2005-12-12, 12:08 #8 akruppa   "Nancy" Aug 2002 Alexandria 246710 Posts Which one is it? Alex   2005-12-12, 18:02 #9 Zeta-Flux   May 2003 7·13·17 Posts It is called "New Solutions of , Mathematics of Computation, Vol. 61, No. 203, p. 361-363. It is available online if you can get to mathscinet. It's a few years old now, and there is a more recent paper giving better bounds, but this one was good enough. Last fiddled with by akruppa on 2005-12-12 at 23:49 Reason: added [tex] tags   2005-12-12, 21:37   #10
fetofs

Aug 2005
Brazil

2×181 Posts Quote:
 Originally Posted by Zeta-Flux It is called "New Solutions of $a^{p-1}\equiv 1 (\operatorname{mod} p^2)$, Mathematics of Computation, Vol. 61, No. 203, p. 361-363.
Could some handy mod out there add the [tex] tags?   2005-12-12, 22:05 #11 Zeta-Flux   May 2003 7×13×17 Posts Ah, so that's how to do it. Last fiddled with by Zeta-Flux on 2005-12-12 at 22:06   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post kurtulmehtap Math 0 2012-09-17 13:04 jnml Puzzles 12 2012-04-28 21:33 Fusion_power Puzzles 14 2008-04-25 11:37 davar55 Puzzles 34 2007-06-12 14:17 maheshexp Math 2 2004-05-29 01:54

All times are UTC. The time now is 07:01.

Mon Apr 12 07:01:12 UTC 2021 up 4 days, 1:42, 1 user, load averages: 2.69, 1.96, 1.68

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.