20180228, 12:10  #1 
Feb 2018
1100000_{2} Posts 
Are Mersenne numbers really square?
What do you think about it ?
JMMA 
20180228, 15:56  #2 
Sep 2003
2×1,291 Posts 
You mean Mersenne numbers with prime exponents.
Nobody knows for sure. If you ever find one that's nonsquarefree, then you automatically also find the third known Wieferich prime, and that would be a pretty big deal. 
20180228, 18:05  #3 
Aug 2006
1011100110010_{2} Posts 
It's an interesting question. I seem to recall a weak consensus that there probably are Mersenne numbers with prime exponents which are not squarefree, but they may be extremely large. Does anyone here have an opinion?

20180228, 18:39  #4 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
(Mod(1,3)*j*p^2+Mod(2,3)*p)*k^2+(Mod(1,3)*j*p+Mod(2,3))*k+Mod(1,3)*j roots mod 3 is what it comes down to partially.

20180228, 20:17  #5  
Feb 2017
Nowhere
F2D_{16} Posts 
Quote:
Any nonsquarefree Mersenne numbers with prime exponent certainly would be "extremely large," at least by my standards. As already pointed out, the exponent would be the next known Wieferich prime. The best known lower bound I could find in a quick online search was A Wieferich Prime Search up to 6.7 × 10^{15}. From the abstract, Quote:


20180228, 20:21  #6 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 

20180228, 21:01  #7  
Aug 2006
2·2,969 Posts 
Quote:
Quote:


20180228, 21:30  #8  
Sep 2003
2·1,291 Posts 
Quote:
There is good reason to believe that all factors of size 64 bits or less of Mersenne exponents under 1 billion have already been found, either by TF or by user TJAOI, as well a considerable percentage of 65 bit factors. So all new Mersenne factors being discovered are in the 10^{20} range, which is beyond the range of the various exhaustive searches that were done by Dorais and Klyve, and I think also by PrimeGrid at one point. So theoretically, every new factor we find has some infinitesimal chance of being the next Wieferich prime. 

20180228, 21:56  #9 
P90 years forever!
Aug 2002
Yeehaw, FL
2^{3}×3×7×43 Posts 

20180228, 23:03  #10  
Sep 2003
2582_{10} Posts 
Quote:
If f is a factor of 2^{p}−1, then 2^{p} ≡ 1 (mod f). In other words, that the modular exponentiation of (2,p,f) equals 1. Modular exponentiation is available in various languages like PHP or Python, and also in the GMP library in the mpz_powm functions. To check for Weiferich primes, you do the exact same thing, except using modulo f squared. If the modular exponentiation of (2,p,f^{2}) ever equals 1, it's time to prepare a press release. From time to time I've run a check over the entire database of factors, it can be done quite quickly. But it would be better to do it each time a factor is reported. 

20180301, 13:07  #11 
"Rashid Naimi"
Oct 2015
Remote to Here/There
3×11×59 Posts 
I think it is not impossible. So looking long enough should yield hits.
The MersennePseudoorimes are a subset of the more general form: (k+1)^pk^p For positive integer k and prime p As such 3^112^11 is divisible by 23^2 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Square Riesel numbers < 3896845303873881175159314620808887046066972469809^2  Citrix  Math  8  20170410 10:50 
Squarefreeness of Mersenne numbers  Qubit  Math  2  20140502 23:51 
square free Mersenne Numbers?  kurtulmehtap  Math  0  20120917 13:04 
Finding the square root of a large mersenne number  Fusion_power  Math  29  20101014 17:05 
Square numbers and binary representation  ET_  Miscellaneous Math  40  20100606 12:55 