20180228, 12:10  #1 
Feb 2018
2^{5}×3 Posts 
Are Mersenne numbers really square?
What do you think about it ?
JMMA 
20180228, 15:56  #2 
Sep 2003
101000010101_{2} 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
2·2,963 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
8369_{10} 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
2×5^{2}×71 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
8,369 Posts 

20180228, 21:01  #7  
Aug 2006
1726_{16} Posts 
Quote:
Quote:


20180228, 21:30  #8  
Sep 2003
29·89 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
3×2,377 Posts 

20180228, 23:03  #10  
Sep 2003
29·89 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
3601_{8} 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 
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