# Talk:Wieferich prime

The following theorem connecting Wieferich primes and Fermat's last theorem was proven by Wieferich in 1909:

Let p be prime, and let x, y, z be integers such that xp + yp + zp = 0. Furthermore, assume that p does not divide the product xyz. Then p is a Wieferich prime.

In 1910, Mirimanoff was able to expand the theorem by showing that, if the preconditions of the theorem hold true for some prime p, then p² must  divide mp − 1-1. 
He proved for m=3. Taro Morishima proved in 1931 for every prime number m not exceeding 31.
—Preceding unsigned comment added by  218.133.184.93 (talk) 14:54, 26 January 2007 


### Added section
Wieferich primes and Fermat numbers.

A Fermat prime cannot also be a Wieferich prime.
W.Banks,F. Luca and I. Shparlinski, “Estimates for Wieferich numbers,” Ramanujan J. 14. (2007), no. 3, 361–378.


### Commentary
This line has been added frequently, with the reference added to the "further reading" section.  This is 

Badly written
Not written with a < ref > tag or {{cite}} template.
Not notable.
Sourced, if at all, to that reference, which may, in turn, not be reliable, notable, or contain the result.
— Arthur Rubin  (talk) 19:28, 26 April 2008 (UTC)

Surely those were References not Further Reading?  Is the Silverman reference used in the text?  Richard Pinch (talk) 07:05, 2 August 2008 (UTC)
I noticed that reference 60 (http://www1.uni-hamburg.de/RRZ/W.Keller/FermatQuotient.html) gives an error 403. HannsEwald (talk) 00:25, 14 October 2014 (UTC)

This section is from 2008. We make new posts at the bottom but I happened to see your edit on my watchlist. The error may be temporary but I have added {{dead link}} with a url in [1] so now there is a working link to an archive. PrimeHunter (talk) 01:41, 14 October 2014 (UTC)
I replaced the link with a link to an archived version of the site in the Wayback Machine. If the site comes back again, a direct link to the site can be added again. -- Toshio Yamaguchi 07:24, 14 October 2014 (UTC)

Is the W. Meissner who found the Wieferich prime 1093 really the physicist Walter Meissner?  Richard Pinch (talk) 07:06, 2 August 2008 (UTC)

I believe this is not the case, and that it was Waldemar Meissner who found the Wieferich prime ([2], [3]). Waldemar Meissner has a very short entry on Czech wiki which basically states "information is missing" --Jvs.cz (talk) 15:08, 5 October 2008 (UTC)
Thanks. I have changed the article to link Waldemar Meissner. PrimeHunter (talk) 19:03, 5 October 2008 (UTC)

Who conjectured that there were only finitely many Wieferich primes?
It seems to me that if you consider 2 to be a random element of 
  
    
      
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          Z
        
        
          /
        
        
          p
          
            2
          
        
        
          Z
        
        
          )
          
            ∗
          
        
      
    
    {\displaystyle (\mathbb {Z} /p^{2}\mathbb {Z} )^{*}}
  
, then 2p − 1 will be 1 mod p2 with probability 1/p, so that the expectation of the number of Wieferich primes less than x would be 
  
    
      
        
          ∑
          
            p
            ≤
            x
          
        
        1
        
          /
        
        p
      
    
    {\displaystyle \sum _{p\leq x}1/p}
  
 which is approximately log(log x), and this grows very slowly but grows to infinity nonetheless.  Experiments along the lines of replacing 2 by various other small numbers supports this hueristic.
Johnny Vogler (talk) 03:16, 15 October 2008 (UTC)

A good point, and Guy's UPINT only says "unknown".  Be bold and change it if you like.   BTW, I moved your new section down here to the bottom as is conventional.  Richard Pinch (talk) 06:03, 15 October 2008 (UTC)
http://primes.utm.edu/glossary/page.php?sort=WieferichPrime says probably finite. PrimeHunter (talk) 11:09, 15 October 2008 (UTC)
If the definition of Wieferich prime to base b is a prime p satisfying bp == b (mod p2) (this definition is completely the same as the original definition if p does not divide b), then if b is a random integer, the property that a given prime p is a Wieferich prime to base b is exactly 1/p. (Besides, if b is a random integer, the property such that there are no Wieferich primes is (1/2) * (1/3) * (1/5) * (1/7) * (1/11) * (1/13) * ... = 0, the same as the property such that there are no primes divide b, since if b is a random integer, the property that a given prime p divides b is also 1/p, the only such b are 1 and -1, thus there may be integers b such that no Wieferich primes base b exist, but the density of the set of these integers b is zero) — Preceding unsigned comment added by 115.82.178.88 (talk) 
Your long parenthentical remark makes no sense. http://primes.utm.edu/glossary/page.php?sort=WieferichPrime has changed since 2008 and now says there are probably infinitely many. This is the common estimate. If a random prime p has estimated chance 1/p of being a Wieferich prime then the expected number of Wiferich primes is the sum of 1/p over all primes p. This sum is infite per Divergence of the sum of the reciprocals of the primes. PrimeHunter (talk) 12:09, 16 June 2018 (UTC)
The sum of inverses of primes is not closed, but in the order of ln(ln(x)).  This corresponds to an indefinite number near \aleph_{-2}.  A given number has an indefinite number of sevenites (Weiferich primes are binary sevenites), but I have not seen any list longer than eighty characters in the thousands of bases I have sampled.  The compound sevenites, such as 7 in base, where at least the cube divides the period, are known not to exist in all bases, although 68 has two (5, 113).
113 in base 68, is the only example of a bond prime with b<p<2400000.  These are bond primes, for if 7 were to have a bond sevenite, the tale would be 0,0,7 in that prime.  Wendy.krieger (talk) 07:46, 6 October 2018 (UTC)

Considering 2 as a random element does not make sense; what is considered random here is 2p-1 modulo p2. More precisely, a heuristic argument here is the following assumption:

2p-1 modulo p2 is equally distributed in the set of (p-1)-th powers 
  
    
      
        {
        
          x
          
            p
            −
            1
          
        
        :
        x
        ∈
        
          
            Z
          
          
            
              p
              
                2
              
            
          
          
            ∗
          
        
        }
      
    
    {\displaystyle \{x^{p-1}:x\in \mathbb {Z} _{p^{2}}^{*}\}}
  
.
This assumption together with 
  
    
      
        
          |
        
        {
        
          x
          
            p
            −
            1
          
        
        :
        x
        ∈
        
          
            Z
          
          
            
              p
              
                2
              
            
          
        
        }
        
          |
        
        =
        p
      
    
    {\displaystyle |\{x^{p-1}:x\in \mathbb {Z} _{p^{2}}\}|=p}
  
 implies that

2p-1 modulo p2 = 1 with the probability 1/p.
Maxal (talk) 01:27, 9 June 2009 (UTC)

"Consider 2p-1 as a random element" of anything is what doesn't make any sense; considering 2 as a random element of the multiplicative group (which is isomorphic to 
  
    
      
        
          
            Z
          
          
            p
          
        
        ×
        
          
            Z
          
          
            p
            −
            1
          
        
      
    
    {\displaystyle \mathbb {Z} _{p}\times \mathbb {Z} _{p-1}}
  
) makes sense, although not quite correct.  One needs to choose 2 as a random element of the multiplicative group with order not dividing p (as those can be be easily seen to be those congruent to 1 modulo p), to get the proper fraction 1/p.  — Arthur Rubin  (talk) 01:33, 9 June 2009 (UTC)
Considering 2p-1 as a random element produces the correct answer, but any good number-theorist can see that it cannot be justified except by 2 being a random element with the precise formulation I now suggest.  — Arthur Rubin  (talk) 01:38, 9 June 2009 (UTC)
First, the group must be 
  
    
      
        
          
            Z
          
          
            
              p
              
                2
              
            
          
          
            ∗
          
        
      
    
    {\displaystyle \mathbb {Z} _{p^{2}}^{*}}
  
 - everything is done modulo p2. Second, it sounds weird - how would you describe distribution of a ″random″ element 2 ? Third, assumption of the uniform distribution (in whatever form) is important here, otherwise there is no way to derive the probability of desired event. Maxal (talk) 01:39, 9 June 2009 (UTC)
btw, ″any good number-theorist″ is not an argument. Maxal (talk) 01:47, 9 June 2009 (UTC)

It's difficult to argue as to which heuristic "make sense", because they're all obviously wrong, in a sense.  Still, it seems the proper formulation is that 2 is a random element of 
  
    
      
        
          
            Z
          
          
            
              p
              
                2
              
            
          
          
            ∗
          
        
      
    
    {\displaystyle \mathbb {Z} _{p^{2}}^{*}}
  
, (not in the subgroup of order p, as 2^p is congruent to 2 modulo p, and hence cannot be congruent to 1 modulo p2) considered in its natural representation as being isomorphic to 
  
    
      
        
          
            Z
          
          
            p
          
        
        ×
        
          
            Z
          
          
            p
            −
            1
          
        
      
    
    {\displaystyle \mathbb {Z} _{p}\times \mathbb {Z} _{p-1}}
  
, which produces the same answer.  Furthermore, "uniform distribution" is essential, but is always implied by "chosen randomly" unless another distribution is stated, and doesn't quite fit the Wikipedia article's definition.
2p-1 can't be random, either.
As for "any good number-theorist" being an argument, probably not.  However, I worked with Erdos on some of his number-theory "puzzles",…  but, I suppose argument from authority isn't a good argument, either, at least in Wikipedia.  — Arthur Rubin  (talk) 01:50, 9 June 2009 (UTC)
The argument that works in Wikipedia is to make your proposal, and see which version gets consensus.  Accuracy is not entirely relevant, no matter how much we would wish otherwise.  May I suggest that you revert the sentence to one of my suggestions after you started editing, per WP:BRD, rather than having the matter disputed.  I believe I'll need to add a dispute tag to the present revision. — Arthur Rubin  (talk) 01:54, 9 June 2009 (UTC)
″it seems the proper formulation is that 2 is a random element″ - prove it, what is a distribution here? Moreover, "proper" or not is an issue of sense. ″I worked with Erdos″ is neither an argument. If you want this kind of argumentation - I have a Ph.D. in computer science - so what? Maxal (talk) 01:55, 9 June 2009 (UTC)
Above I explained precisely what is motivation of my changes, while you did not provide any arguments besides ″any good number-theorist can see″ and ″I worked with Erdos″. Maxal (talk) 01:59, 9 June 2009 (UTC)

You explained what your motivation was, and my argument is:
It doesn't make any sense.
That's subjective, but I think the proper thing to do until we reach a Wikipedia consensus is either to revert that line to the status before you started editing, or to remove the rationale entirely.  I cannot do the former, because of WP:3RR, but the latter seems to make more sense until we can find a reference.  — Arthur Rubin  (talk) 02:05, 9 June 2009 (UTC)
OK. Let's cool down. The argument ″it does not make sense″ was exactly mine with respect to treating 2 as a random element. I still do not see what exactly that is supposed to mean. 
What I suggested may not be complete either but I will try to formalize it further later on. Maxal (talk) 02:45, 9 June 2009 (UTC)
It seems to be simpler to use a somewhat reverse assumption - that the (p-1)-th degree roots of unity modulo p2 behave as uniformly distributed random elements in 
  
    
      
        
          
            Z
          
          
            
              p
              
                2
              
            
          
          
            ∗
          
        
      
    
    {\displaystyle \mathbb {Z} _{p^{2}}^{*}}
  
. Since the total number of these roots is p-1, the probability that 2 is one of them is

  
    
      
        (
        p
        −
        1
        )
        
          /
        
        
          |
        
        
          
            Z
          
          
            
              p
              
                2
              
            
          
          
            ∗
          
        
        
          |
        
        =
        (
        p
        −
        1
        )
        
          /
        
        (
        p
        (
        p
        −
        1
        )
        )
        =
        1
        
          /
        
        p
      
    
    {\displaystyle (p-1)/|\mathbb {Z} _{p^{2}}^{*}|=(p-1)/(p(p-1))=1/p}
  

Maxal (talk) 03:18, 9 June 2009 (UTC)
The argument that the Fermat quotient of p should be divisible by p with probability 1/p, leading to the heuristic estimate that the number of Wieferich primes below a given limit n should be asymptotic to log log n, appears in Andrew Granville, "Some Conjectures related to Fermat's Last Theorem," in Richard A. Mollin, ed., Number Theory (1990), pp. 177-92, at p. 178; available online at http://www.dms.umontreal.ca/~andrew/PDF/ConjFLT.pdf.
John Blythe Dobson (talk) 21:01, 9 March 2011 (UTC)

Also, Paulo Ribenboim, in My Numbers, My Friends (Springer, 2000), 226, writes: "Since nothing to the contrary is known, it may be assumed (heuristically) that for each prime p the probability that (2^p - 1)/p = 0 (mod p) is just 1/p since there are p residue classes modulo p." He then gives the same asymptotic estimate as Granville's. John Blythe Dobson (talk) 19:34, 18 March 2011 (UTC)

At the beginning of the article, a Wieferich prime is defined as a prime p such that p2 divides 2p-1-1. This is correct but I think most sources use to define a Wieferich prime as a prime number satisfying the congruence 2p-1 ≡ 1 (mod p2) (see for example [4]).
Thus I propose changing the first sentence of the lead seaction
from
In number theory, a Wieferich prime is a prime number p such that p2 divides 2p−1−1; compare this with Fermat's little theorem, which states that every odd prime p divides 2p−1−1.
to
In number theory, a Wieferich prime is a prime p satisfying the congruence 2p-1 ≡ 1 (mod p2); compare this with Fermat's little theorem, which states that every odd prime satisfies the congruence 2p-1 ≡ 1 (mod p).
Toshio Yamaguchi (talk) 18:20, 3 July 2010 (UTC)

First, saying that p2 divides 2p-1-1 is equivalent to saying that 2p-1 ≡ 1 (mod p2). Second, the former definition does not require knowledge of congruences and their notation. Therefore, I'm opposite to changing the current definition. Maxal (talk) 14:34, 4 July 2010 (UTC)
Ok, sure you're right with this, but just let me explain what my intention is. I would like to add a section about the Near-Wieferich primes to the article and a list of all known examples. As Near-Wieferich primes are commonly defined by the congruence  2(p-1)/2 ≡ ±1 + Ap (mod p2), it would be easier to understand the connection between Wieferich and Near-Wieferich primes. Toshio Yamaguchi (talk) 15:16, 5 July 2010 (UTC)

I don't see problems with "understanding the connection" here. Moreover, congruences are already used in the article. It's just not a good idea to use them in the header, which should give as much non-technical introduction of the subject as possible. Maxal (talk) 15:54, 5 July 2010 (UTC)
I added the section about Near-Wieferich primes. Toshio Yamaguchi (talk) 19:54, 5 July 2010 (UTC)
The citations given are not fully consistent with the citation style used in the rest of the article, as I had some problems generating those reflist citations. Maybe I'm a bit stupid, but can someone explain how I get a citation with these small numbers. As soon as I got this I will change the citations for consistency's sake. Toshio Yamaguchi (talk) 21:04, 5 July 2010 (UTC)

I've edit the first of your citations. Please turn the second into a reference as well. Maxal (talk) 22:50, 5 July 2010 (UTC)
Done. Toshio Yamaguchi (talk) 00:51, 6 July 2010 (UTC)
I removed the reference to the Dorais&Klyve paper in the Near-Wieferich prime section, as they seem to use a modified version of the Near-Wieferich congruence in their paper and thus seem to get other Near-Wieferich prime values than the other searches. Instead I used the paper by Crandall, Dilcher and Pomerance as well as the paper by Knauer and Richstein as reference, as they seem to use the unaltered definition, which is also used in the current search by Wieferich@Home. Toshio Yamaguchi (talk) 09:41, 6 July 2010 (UTC)

The section about the connection between Wieferich primes and Fermat's last theorem currently says:
"The following theorem connecting Wieferich primes and Fermat's last theorem was proven by Wieferich in 1909:

Let p be prime, and let x, y, z be integers such that xp + yp + zp = 0. Furthermore, assume that p does not divide the product xyz. Then p is a Wieferich prime.
In 1910, Mirimanoff was able to expand the theorem by showing that, if the preconditions of the theorem hold true for some prime p, then p2 must also divide 3p − 1 − 1."
This reads as if when a prime p satisfies

  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  

then it must also satisfy

  
    
      
        
          3
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
.
This however is not true as 1093 and 3511 only satisfy

  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  

and 11 and 1006003 only satisfy

  
    
      
        
          3
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  

but none of these primes satisfies both congruences. Am I missing something here? Toshio Yamaguchi (talk) 18:40, 15 March 2011 (UTC)

The preconditions of the theorem here are
"Let p be prime, and let x, y, z be integers such that xp + yp + zp = 0. Furthermore, assume that p does not divide the product xyz."
and this is not the same as 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
. In other words, if p satisfies the preconditions, then 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 as well as 
  
    
      
        
          3
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
, but this does not work in opposite direction. Maxal (talk) 21:24, 16 March 2011 (UTC)
I think it actually should be
If p satisfies the preconditions, then 
  
    
      
        
          a
          
            p
            −
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        ≡
        1
        
          
          (
          mod
          
          
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    {\displaystyle a^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 for some 
  
    
      
        a
        
        
      
    
    {\displaystyle a\,\!}
  
. This does not necessarily mean 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
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    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 and also 
  
    
      
        
          3
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
. It means 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 or maybe 
  
    
      
        
          3
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 or maybe 
  
    
      
        
          5
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 5^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 or if none of these any 
  
    
      
        
          a
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle a^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 for larger a. Thus your statement "then 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 as well as 
  
    
      
        
          3
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
" is incorrect, since for example 1093 only satisfies 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
 but not also 
  
    
      
        
          3
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
        
        
      
    
    {\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}\,\!}
  
. Toshio Yamaguchi (talk) 23:20, 16 March 2011 (UTC)
No, it's correct. 1093 does not satisfy the preconditions, since there are no such x, y, z that x1093 + y1093 + z1093 = 0 with 1093 not dividing xyz. Thus neither Wieferich theorem, nor Mirimanoff theorem is applicable to 1093. In other words, p = 1093 just happens to satisfy 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}}
  
 but it does not satisfy the preconditions of Wieferich theorem. Maxal (talk) 12:50, 17 March 2011 (UTC)
I think I now have understood it. Wieferich simply showed if there were a prime p, such that xp + yp + zp = 0, then this prime p would have to satisfy 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}}
  
. 1093 simply satisfies 
  
    
      
        
          2
          
            p
            −
            1
          
        
        ≡
        1
        
          
          (
          mod
          
          
            p
            
              2
            
          
          )
        
      
    
    {\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}}
  
 without there being integers x, y, z such that xp + yp + zp = 0. Obviously Andrew Wiles showed with is proof of Fermat's last theorem, that the Wieferich case is impossible. Toshio Yamaguchi (talk) 17:31, 19 March 2011 (UTC)
Right, and the interesting is the converse formulation: for all primes p which do not satisfy the Wieferich-condition FLT is automatically true. And after this we need only test that extremely rare primes p for which the Wieferich-condition is true - so for all p<10^12 or so we need only test 1093 end 3511. Nice, huh? Gotti 21:28, 24 March 2011 (UTC)  — Preceding unsigned comment added by Druseltal2005 (talk • contribs) 

I added a table containing all known near-Wieferich primes to the article a while ago. The reason for this is, that most of the recent searches for Wieferich primes also spent a considerable amount of computing power for searching near-Wieferich primes. Furthermore, the Wieferich primes can be considered special cases of near-Wieferich primes. However, some of the results regarding near-Wieferich primes are unpublished. The latest published result is that by Knauer and Richstein. The results of all further search efforts are unpublished as of 2011. More specifically the results of the search to 3x1015 by P. Carlisle, R. Crandall and M. Rodenkirch are unpublished (although the search is mentioned in the given reference). Furthermore, the results of the search to 6.7x1015 by Dorais and Klyve are also unpublished. The near-Wieferich primes found by Dorais and Klyve are from the unpublished draft that can be downloaded from their website. The near-Wieferich primes in [1.25x1015, 3x1015] were also added by me, after I received the values from Mark Rodenkirch via email. I would like to hear the opinions of other editors about what to do with the near-Wieferich primes in this article. Toshio Yamaguchi (talk) 12:06, 27 March 2011 (UTC)

I think the table seen by pressing show at Wieferich prime#Generalizations is a little long. Do you have a source for "considerable amount of computing power for searching near-Wieferich primes"? As far as I know, nobody spends non-trivial time searching for near-Wieferich primes. They are searching for Wieferich primes and I think the best known algorithm requires a computation of the A value in the table, or something equivalent to it (2(p-1)/2 modulo p2). Besides testing whether A is 0 (which would be a Wieferich prime), once A has been computed it is trivial to test whether abs(A) happens to be small, called a near-Wieferich prime. One of the purposes of publishing near-Wieferich primes is to at least publish something (no Wieferich prime has been found since 1922), and give some empirical evidence that your search for Wieferich primes was likely to be exhaustive and correct. The expected number of near-Wieferich primes in an interval can be computed. If an alleged Wieferich prime search found much fewer near-Wieferich primes than expected then it's a sign that the search was probably incorrect and others may have to search the same interval. PrimeHunter (talk) 00:14, 28 March 2011 (UTC)
I don't have a source for "considerable amount of computing power for searching near-Wieferich primes". It is just a guess. In order to test if a given prime p is a near-Wieferich prime, one has to test whether p satisfies 2(p−1)/2 ≡ ±1 + Ap (mod p2) for all A in [-100, 100]. That would mean testing each p with 201 individual congruences. Thus the search for Wieferich primes, where A = 0 would consume only 
  
    
      
        
          
            1
            201
          
        
      
    
    {\displaystyle {\frac {1}{201}}}
  
 of the total processing power spent on a single p, while the other 
  
    
      
        
          
            200
            201
          
        
      
    
    {\displaystyle {\frac {200}{201}}}
  
 of the power are spent on testing p for being a near-Wieferich prime. However this is just my guess and I don't have a source for this. The length of the table is the reason, why I used a collapsible table for presenting the near-Wieferich primes. The question is, what encyclopedic value the presentation of the near-Wieferich values in this article serves. Whoever is interested in the values could consult the given sources. Thus I would not oppose a decision to remove the two tables presenting the values of the near-Wieferich primes if there is consensus to do so. Toshio Yamaguchi (talk) 10:35, 28 March 2011 (UTC)
My guess it that people spend more than 99% of the time computing rp = 2(p-1)/2 modulo p2, and they would have done the same if they only searched Wieferich primes. After that they spend less than 1% computing A = (rp ± 1)/p, and testing abs(A) ≤ 100. I suggest we show an uncollapsed table of abs(A) ≤ 10. PrimeHunter (talk) 13:07, 28 March 2011 (UTC)
Right, it really should suffice to compute 2(p-1)/2 only once for each p and then simply test it mod p2 two times (once for +1 and once for -1) with all 201 permitted A values for each p. The most resource consuming part would thus in fact be computing 2(p-1)/2 which requires ((p-1)/2)-1 multiplication operations. Your suggestion for the table sounds good too. Toshio Yamaguchi (talk) 14:09, 28 March 2011 (UTC)
I reduced the lenght of both tables of near-Wieferich primes and made both uncollapsible. Toshio Yamaguchi (talk) 22:04, 28 March 2011 (UTC)

It looks good to me. PrimeHunter (talk) 22:19, 28 March 2011 (UTC)
The table became incomplete. Basically, you reduced the table that was originally based on small values of 
  
    
      
        
          |
          
            
              
                ω
                (
                p
                )
              
              p
            
          
          |
        
      
    
    {\displaystyle \left|{\frac {\omega (p)}{p}}\right|}
  
, using the value of 
  
    
      
        
          |
        
        ω
        (
        p
        )
        
          |
        
      
    
    {\displaystyle |\omega (p)|}
  
, which does not make much sense. In particular, p = 46262476201 would need formally be in this table since 
  
    
      
        
          |
        
        ω
        (
        p
        )
        
          |
        
        =
        10
      
    
    {\displaystyle |\omega (p)|=10}
  
 (despite that the value of 
  
    
      
        
          |
          
            
              
                ω
                (
                p
                )
              
              p
            
          
          |
        
      
    
    {\displaystyle \left|{\frac {\omega (p)}{p}}\right|}
  
 is not small enough). Instead, of selecting values based on 
  
    
      
        
          |
          
            ω
            (
            p
            )
          
          |
        
        ≤
        100
      
    
    {\displaystyle \left|\omega (p)\right|\leq 100}
  
, I suggest to select them based on 
  
    
      
        
          |
          
            
              
                ω
                (
                p
                )
              
              p
            
          
          |
        
        ≤
        q
        ×
        
          10
          
            −
            14
          
        
      
    
    {\displaystyle \left|{\frac {\omega (p)}{p}}\right|\leq q\times 10^{-14}}
  
 for some q (e.g., q = 1) that would be consistent with the original selection criterion. Maxal (talk) 22:51, 29 March 2011 (UTC)
I changed the second near-Wieferich table according to your suggestion. Toshio Yamaguchi (talk) 07:44, 30 March 2011 (UTC)
The first table of near-Wieferich primes is either incomplete or incorrectly described. For example, why primes 3 or 867457663 are not shown there? In particular, there are a lot of near-Wieferich (as they are defined) in the interval (3511,46262476201), i.e., between the second and third entries, but none are shown. Maxal (talk) 15:20, 23 December 2011 (UTC)
I think I made a mistake in the description. I didn't list any near-Wieferich primes in [1, 1×109] other than 1093 and 3511 since I don't have their values. I took the values from the following sources:
http://www.loria.fr/~zimmerma/records/Wieferich.status (the values are identical to those from Crandall et al and Knauer and Richstein)
Crandall, Dilcher and Pomerance
Knauer and Richstein
the last value is from personal communication
I don't know why I omitted for example 3520624567, which would need to be in the table under the current definition, but have to admit that it might be the result of sloppy work on my part. Toshio Yamaguchi (talk) 19:30, 23 December 2011 (UTC)
Comment: See below Proposed merger from List of near Wieferich primes to this article. Northamerica1000(talk) 22:55, 28 December 2011 (UTC)

Maxal, I reverted your last edits, because you changed the book sources for the search by Carlisle, Crandall and Rodenkirch to point to the english editions of the books. Please note that I assume good faith on your part, but the english and german editions of the books seem to differ from each other. The Carlisle, Crandall and Rodenkirch search is only mentioned in the german editions of those books. Thus the english editions do not verify the given information, whereas the german editions do. Btw the books only mention this search, but do not give the values of the near-Wieferich primes found. I received these values from Mark Rodenkirch on request via email. The last near-Wieferich prime in the current table is one of those values and they are not published in the books or anywhere else. Toshio Yamaguchi (talk) 10:27, 29 March 2011 (UTC)

Actually, you are correct about the Carlisle, Crandall and Rodenkirch search—it's mentioned only in German edition of "The little book of bigger primes" ("Die Welt der Primzahlen: Geheimnisse und Rekorde"). So, I'm retaining it. But for "My numbers, my friends: popular lectures on number theory", it is pointless to refer to the German edition. The corresponding pages are missing at Google/Books but that does not disqualify the book as a source of information. Moreover, this is English Wikipedia and English sources are preferable for it than sources in any other language. Partially reverting back. Maxal (talk) 16:41, 29 March 2011 (UTC)

For a cyclotomic generalization of the Wieferich property:(np − 1)/(n − 1) divisible by q2, there are solutions like

(35 − 1)/(3 − 1) = 112
and even with exponents higher than 2, like in
(196 − 1)/(19 − 1)  ≡ 0 (mod 73).
I removed the above statement from the article, since I was unable to find a source mentioning this kind of generalization of the Wieferich property. It has been placed here for further discussion. If anyone is able to provide sources for this, it can be pasted back into the article from here with a source. Toshio Yamaguchi (talk) 18:15, 14 April 2011 (UTC)

I can not find a reference either to a 'general wieferich prime'.  I made a post at dozensonline / mathematics / number theory / sevenites in September 2012 about this.  I have been studying this for many years without finding a general name, so i use 'sevenite' (after 7 in base 18), for the general 1/p distribution.  There seems to be a quite a number of wikipedia edits after 2015 when a user found that the binary sevenites are Wieferich, the trinary sevenites are Marianoff, the decimal sevenites are Shank's.  David Well's "curious and interesting numbers" mentions examples of five sevenites, but only the entry for 1093 carry Wieferich's name.  E. Dickson's "History of the Theory of Numbers" has an unnamed twelve-page chapter four on the subject, but mentions Wieferich only in a five-line paragraph on the nineth page.  Using the name Wieferich is a bad proposition, since the research stretches nearly 200 years before his paper, and his paper has nothing to do with the general divisibility of p^n | a^{p-1}-1. Wendy.krieger (talk) 10:14, 14 April 2018 (UTC)

Nathanson, M. B. (2000). Elementary methods in number theory. Springer. p. 187. ISBN 0-387-98912-9.
This source calls a prime number p satisfying 2p − 1 ≢ 1 (mod p2) a Wieferich prime, while it seems that most other sources agree that a Wieferich prime is a prime satisfying 2p − 1 ≡ 1 (mod p2).
Should there be a note about this in the article? Toshio Yamaguchi (talk) 18:22, 9 June 2011 (UTC)

It is amazing that he even gives "examples" - he should have found out at least then, that he made something very wrong here... --Gotti 11:34, 14 June 2011 (UTC)  — Preceding unsigned comment added by Druseltal2005 (talk • contribs) 
Yeah, I think I will add the source to Further reading with a note pointing out this inconsistency. A reader searching Google books might stumble over this and after reading our article here and other sources, this might cause confusion to the reader. Toshio Yamaguchi (talk) 12:35, 14 June 2011 (UTC)

In the section about the connection with the Mersenne and Fermat primes it says
"A prime divisor p of Mq, where q is prime, is a Wieferich prime if and only if p2 divides Mq."
Furthermore it says
"Thus, a Mersenne prime cannot also be a Wieferich prime"
I do not see how the second statement follows from the first one. Unless someone can provide a source connecting these two statements in that way, I am going to remove the second statement, as this looks like original research. Toshio Yamaguchi (talk) 13:24, 29 November 2011 (UTC)

A Mersenne prime is a prime divisor of itself. If Mq is prime then p=Mq is a prime divisor of Mq, but p2 = Mq2 does clearly not divide Mq. PrimeHunter (talk) 14:45, 29 November 2011 (UTC)
Okay, I see it now. Thanks for the explanation. Toshio Yamaguchi (talk) 15:09, 29 November 2011 (UTC)

A non-mathematician writes... The lead says that they were "first described" in 1909, at which time the theorems were "already well known." I'm sure this makes sense to mathematicians, but it reads a bit oddly to the non-specialist. Tigerboy1966 (talk) 11:59, 7 December 2011 (UTC)

The theorems referred to are Fermat's last theorem and Fermat's little theorem. Both of those theorems had already been studied earlier by mathematicians, which is confirmed by citations 4 and 5 respectively, so I thought that would be clear. Toshio Yamaguchi (talk) 12:06, 7 December 2011 (UTC)
I clarified it in this edit. I hope that's better. Thanks for pointing that out. Toshio Yamaguchi (talk) 12:17, 7 December 2011 (UTC)
Thanks. Also the new headings instead of "Other Properties" are good. Tigerboy1966 (talk) 12:27, 7 December 2011 (UTC)
There is a rather good discussion in Dickson's "History of the theory of Numbers" (1919).  Chapter 4 vol 1 is entirely devoted to the history of this: it goes back as far as Euler.  The chapter runs to eight pages in the Dover reprint.  Wieferich is mentioned on page 110, the sixth page, in a five-line paragraph.  I have been playing around with these for nearly forty years, and keeping an eye out for any general term.  (I use 'sevenite' after the example of 7 in base 18).
Yates (Repunits) lists the sevenites for odd primes, and notes the three known decimal ones.  Wells (Curious and Interesting Numbers) lists five, but attaches the name of Wieferich only to 1093.  The Primes Pages from 2006 do not list any general class name for sevenites.
I only heard of 'Wieferich prime' in the proposed sense after I introduced the term 'sevenite' onto DozensOnline in 2012 or so.  I counceld against calling then "Wieferich Primes", since Wieferich had very little to do with them, and using this term would lead students astray.  I had the same problem with 'Wythoff Notation' (which has nothing to do with Wythoff, but relates to an honour name bestowed on a symbol that modifies the Schwarz-triangle notation).
I considered the term disruptive to science, and we ought find one that does not mislead researchers Wendy.krieger (talk) 11:08, 5 October 2018 (UTC)

Merging the contents from the List of near Wieferich primes article would benefit this article, by making it more complete, and hence, more encyclopedic. After a merger, the addition of a collapsed table would address the length of the finalized table. Northamerica1000(talk) 22:53, 28 December 2011 (UTC)

I don't think the complete list should be in this article. A list of all near Wieferich primes in [1×109, 3×1015] can be found in OEIS as A195988. Toshio Yamaguchi (talk) 08:49, 29 December 2011 (UTC)
This was exactly my argument. Merging does not make sense as all essential information is already present in Wieferich prime article. The article List of near Wieferich primes should be simply deleted. Maxal (talk) 01:10, 30 December 2011 (UTC)
I nominated the list article for deletion (see Wikipedia:Articles for deletion/List of near Wieferich primes). Toshio Yamaguchi (talk) 01:28, 30 December 2011 (UTC)

In the section 'Connection with Mersenne and Fermat primes' I included the following statement:
It was observed that M1092 is divisible by 10932 and M3510 is divisible by 35112.
I did this, because http://www.elmath.org/index.php?id=display_subject&subject=2 says this is a "remarkable" fact. I don't have access to the paper by Guy which Miroslav Kures cites on the Wieferich@Home project homepage, but thinking about this again I believe this observation is rather trivial. Given that for a Wieferich prime qp(2) ≡ 0 (mod p) and the numerator of the Fermat quotient is always Mp-1 it follows as a corollary that for a Wieferich prime p2 divides Mp-1. Thus I believe the statement should perhaps be removed. Opinions? Toshio Yamaguchi (talk) 15:41, 8 January 2012 (UTC)

A agree that this statement is trivial and placed out of the context. So I support its removal. Maxal (talk) 16:27, 8 January 2012 (UTC)
I removed it. Toshio Yamaguchi (talk) 21:33, 9 January 2012 (UTC)

This review is transcluded from Talk:Wieferich prime/GA1. The edit link for this section can be used to add comments to the review.
Reviewer: GreatOrangePumpkin (talk · contribs) 10:46, 7 May 2012 (UTC)

GA review (see here for what the criteria are, and here for what they are not)
It is reasonably well written.
a (prose):  b (MoS for lead, layout, word choice, fiction, and lists): 
What does "≡" means? If and only if? And "∤" means "restriction to"?
"≡" in the context of this article denotes a congruence on the integers. -- Toshio Yamaguchi (tlk−ctb) 18:47, 8 May 2012 (UTC)
"∤" here means "does not divide" (it is listed at List of mathematical symbols#Symbols). -- Toshio Yamaguchi (tlk−ctb) 18:47, 8 May 2012 (UTC)
I wikilinked the symbol "∤" to an article where it is explained. -- Toshio Yamaguchi (tlk−ctb) 18:32, 9 May 2012 (UTC)
"log log x" a word too much or intentional?
This is intentional. It denotes the logarithm of the logarithm of x. I added parentheses for clarity here. -- Toshio Yamaguchi (tlk−ctb) 18:47, 8 May 2012 (UTC)
Similar uncommon units should be linked
What in particular do you think should be linked? -- Toshio Yamaguchi (tlk−ctb) 18:51, 9 May 2012 (UTC)
Can you explain what does the small omega in the second table refers to? If you do this I am happy to pass this article.
Explained in this edit. -- Toshio Yamaguchi (tlk−ctb) 20:21, 11 May 2012 (UTC)
Red links should be created or explained in the article (for example, I don't know what a "Catalan pseudoprime" is)
Redlinks First case of Fermat's last theorem and Catalan pseudoprime eliminated by creating those articles. -- Toshio Yamaguchi (tlk−ctb) 20:51, 9 May 2012 (UTC)
Non-english sources should be labelled as such.
Done. -- Toshio Yamaguchi (tlk−ctb) 19:25, 9 May 2012 (UTC)
Suggest to remove the red links in the reference section, as it is unlikely they will be created.
Done. -- Toshio Yamaguchi (tlk−ctb) 19:46, 9 May 2012 (UTC)
It is factually accurate and verifiable.
a (references):  b (citations to reliable sources):  c (OR): 
Overall ok
It is broad in its coverage.
a (major aspects):  b (focused): 

It follows the neutral point of view policy.
Fair representation without bias: 

It is stable.
No edit wars, etc.: 

It is illustrated by images, where possible and appropriate.
a (images are tagged and non-free images have fair use rationales):  b (appropriate use with suitable captions): 

Overall: On hold to give you time fixing the problems.--GoPTCN 11:43, 8 May 2012 (UTC) Pass
Pass/Fail: 

(Note): My common practice is to paste the template into the GAN. I often check points 5 and 6 (stability and pictures) before reviewing because they are easy to check. "Neutral" is confusing, as it may also mean that I did not review this points. Ok, now I will resume my review :)--GoPTCN 11:14, 8 May 2012 (UTC)

I created a navbox template for the listing of unsolved problems (see User:Toshio Yamaguchi/Template:Unsolved 2) and propose to remove the current instance of Template:Unsolved at Wieferich prime#History and search status and replace it with this navbox (after having been moved into template namespace). See my sandbox for how the template looks like. Do other editors agree with this step? -- Toshio Yamaguchi (tlk−ctb) 21:51, 2 June 2012 (UTC)

I disagree.  Navboxes are to be used for navigation, not for explanations per WP:NAVBOX.  I don't particularly like {{unsolved}} either, but it has passed 4 deletion discussions already. Ryan Vesey Review me! 23:57, 2 June 2012 (UTC)
Then perhaps these unsolved problems should just be incorporated into the article text and the template be removed entirely. Just because it has passed 4 TfDs is not an endorsement of the templates presence in this article. -- Toshio Yamaguchi (tlk−ctb) 07:19, 3 June 2012 (UTC)
I think that would be the best solution. Ryan Vesey Review me! 07:27, 3 June 2012 (UTC)

Wieferich_prime#Equivalent_congruences states that

2kp = 2 (mod p2) for all k.
But if k=1, you have

2p = 2 (mod p2)
which is true by definition. Squaring it yields

22p = 4 (mod p2)
which is clearly not 2.
The best I've come up with yet is

2p * 2(p-1)k = 2 * 1 (mod p2) for all k,
and this means that k=p gives the result

2p * 2(p-1)p = 2 * 1 (mod p2),
which simplifies to

2p^2 = 2 (mod p2). - ARGH!!!
I tried to prove you wrong, but that's actually the result which CAN be found here. Can anyone please clarify the steps in that sentence, Thus a Wieferich prime satisfies (...) for all integers k ≥ 1., and tell me where I did wrong, I feel my neurons melt when thinking about it any longer. Never was my signature any truer, I'm afraid. - ¡Ouch! (hurt me / more pain) 11:07, 7 September 2012 (UTC)

As the article congruence relation#Basic example states congruence modulo an integer n is compatible with addition and multiplication. For any intgers m and n the statement m ≡ m (mod n) is always true. If we consider three integers a, b and n for which a ≡ a (mod n) and b ≡ b (mod n) hold, then also does ab ≡ ab (mod n). So for a congruence if we multiply both sides by the same integer, the congruence must always hold. Now consider the Wieferich congruence 2p-1 ≡ 1 (mod p2), which here holds for p. Thus we can again multiply both sides with the same integer, say 2 and get 2p ≡ 2 (mod p2).
And that's the point where it gets all wrong... - ouch
Repeating this we get 2p+1 ≡ 2 (mod p2), 2p+2 ≡ 2 (mod p2), 2p+3 ≡ 2 (mod p2), etc. We can simply repeat this until we get 2p+p ≡ 2 (mod p2), which is 22p ≡ 2 (mod p2). Likewise, we can repeat it until we get 23p ≡ 2 (mod p2), 24p ≡ 2 (mod p2) and generally 2sp ≡ 2 (mod p2) for any integer s. Now we set s=p and
I suppose that's the point what made the error hard to find, after some wrong steps, the whole thing seems to add up to the correct conclusion. - ouch
get 2p2 ≡ 2 (mod p2). -- Toshio Yamaguchi (tlk−ctb) 11:57, 7 September 2012 (UTC)

Huh ? Yes, we have 2p ≡ 2 (mod p2) for a Wieferich prime, but if we square both sides we get 22p ≡ 4 (mod p2), so it certainly not true that 2kp ≡ 2 (mod p2) for all k ≥ 1. However, we can argue as follows:-

  
    
      
        
          2
          
            p
          
        
        ≡
        2
        
        mod
        
        
        
          p
          
            2
          
        
        ⇒
        
          2
          
            
              p
              
                2
              
            
          
        
        =
        (
        
          2
          
            p
          
        
        
          )
          
            p
          
        
        ≡
        
          2
          
            p
          
        
        ≡
        2
        
        mod
        
        
        
          p
          
            2
          
        
      
    
    {\displaystyle 2^{p}\equiv 2\mod p^{2}\Rightarrow 2^{p^{2}}=(2^{p})^{p}\equiv 2^{p}\equiv 2\mod p^{2}}
  

and so, by induction,

  
    
      
        
          2
          
            
              p
              
                k
              
            
          
        
        ≡
        2
        
        mod
        
        
        
          p
          
            2
          
        
         
        ∀
         
        k
        ≥
        1
      
    
    {\displaystyle 2^{p^{k}}\equiv 2\mod p^{2}\ \forall \ k\geq 1}
  

Gandalf61 (talk)
Hmm, yes, if we multiply 2 by an integer greater than 1, the result will no longer be equal to 2. I really should take more care before spitting out stuff like that.... -- Toshio Yamaguchi (tlk−ctb) 20:51, 7 September 2012 (UTC)
(Sorry Toshio for butchering your reply, but it took me 5 minutes to spot the error, and IMO this is one of the rare cases where editing you makes the issue easier to understand for third parties. I didn't change your words, but pointed out which equations are off. Not meant as humiliation, but to save other readers 5 minutes of their time.)
I found an approach which proves above equation without damaging my synapses. It goes as follows:
I take the following equations,

Eq1: 2p-1 = 1 (mod p2), which holds for any Wieferich prime, and
Eq2: 21 = 2 (mod p2) (duh!).
Once I realized that pn - 1 is a multiple of (p-1), all

2p^n - 1 = 1 (mod p2),
because they are powers of 2p-1.
Now multiply by Eq2 to get the result,

2p^n = 2 (mod p2). This will hold for all integers n>1.
Maybe this could go into the article, as it is easier to grasp IMO than what is in there now. Comments? - ¡Ouch! (hurt me / more pain) 08:14, 10 September 2012 (UTC)
It's not true in general.  bp^2-p = 1, (mod p2) is the general case.  For example 2^25 = 33554432, is equal to 7, mod 25, not 2.  Likewise, 3^9 = 512 which is 8, mod 9, not the value 2. By fermat's little theorm, b^(p-1) = 1, modulo p, so we get b^n(p-1)+1 = b, modulo p.  For all primes, b^(p^2-p) = 1, mod p² Wendy.krieger (talk) 11:15, 5 October 2018 (UTC)

Oops. I think the "Equivalent congruences" section is wrong again..

"Raising both sides of the congruence to the power p shows that a Wieferich prime also satisfies
2p  2    ≡ 4 (mod p2)"
I get 2p2 = (2p)p ≡ 2p (mod p2) instead.
Which reduces to 2. Induction yields 2pk ≡ 2 (mod p2) for all positive integers k. - ¡Ouch! (hurt me / more pain) 09:29, 14 September 2012 (UTC)

As we found earlier, multiplying the right side of the congruence by two changes the value there, so it cannot be true that 2pk ≡ 2 (mod p2) for all positive integers k. -- Toshio Yamaguchi (tlk−ctb) 09:38, 14 September 2012 (UTC) Nonsense on my part. -- Toshio Yamaguchi (tlk−ctb) 11:51, 17 September 2012 (UTC)
Then tell me, where did I multiply by two? - ¡Ouch! (hurt me / more pain) 07:44, 17 September 2012 (UTC)
It seems to me the correct statement would be that 2pk ≡ 2k (mod p2) for all positive integers k. -- Toshio Yamaguchi (tlk−ctb) 09:54, 14 September 2012 (UTC)
2pk = (2p)pk - 1, not (2p)k.
OMG... - ¡Ouch! (hurt me / more pain) 07:44, 17 September 2012 (UTC)
The fact that pk - pk - 1 is a multiple of (p - 1) proves that p, p2, ... , pk all reduce to 1 (modulo (p - 1)).
Which in turn proves that all 2pkare 
2 times a power of (2p - 1 ) (mod p2),
the latter factor being 1 by definition.
I hope. {Headache} - ¡Ouch! (hurt me / more pain) 09:29, 14 September 2012 (UTC)

Toshio - I have reverted you changes, which were wrong. See the section above for two proofs that 2pk ≡ 2 (mod p2 for all positive integers k as long as p is a Wieferich prime. Or try some numerical examples if you don't follow the proof. For example:

  
    
      
        
          2
          
            
              1093
              
                2
              
            
          
        
        ≡
        2
        
        mod
        
        
        
          1093
          
            2
          
        
      
    
    {\displaystyle 2^{1093^{2}}\equiv 2\mod 1093^{2}}
  

Gandalf61 (talk) 09:55, 14 September 2012 (UTC)
I think I have been mixing up some things. Yes, the residue of 21093 mod 10932 is 2 and so 210932 is congruent to 2 modulo 10932. -- Toshio Yamaguchi (tlk−ctb) 10:15, 14 September 2012 (UTC)

Somewhere in the article should be the factorization of 1092 and 3510. The question is where? - Virginia-American (talk) 10:41, 25 September 2012 (UTC)

I would suggest rename section 'Binary periodicity of p-1' with 'Properties of p-1' and show factorizations there. Maxal (talk) 01:12, 26 September 2012 (UTC)
I don't think the prime factorizations of those two numbers are particularly interesting, as I guess that's what is supposed to be added. More interesting (in my opinion) are the divisors of those two numbers (see the section An unexpected property of the divisor-sums here). -- Toshio Yamaguchi (tlk−ctb) 13:17, 27 September 2012 (UTC)

When I search the interval [1000, 1100] with wwww and tell it to report special instances with |A| ≤ 1000 I get
1009 is a special instance (+1 +296 p)
1009 is a special instance (+1 -713 p)
1013 is a special instance (-1 +41 p)
1013 is a special instance (-1 -972 p)
1019 is a special instance (-1 +657 p)
1019 is a special instance (-1 -362 p)
1021 is a special instance (-1 +644 p)
1021 is a special instance (-1 -377 p)
1031 is a special instance (+1 +318 p)
1031 is a special instance (+1 -713 p)
1033 is a special instance (+1 +251 p)
1033 is a special instance (+1 -782 p)
1039 is a special instance (+1 +872 p)
1039 is a special instance (+1 -167 p)
1049 is a special instance (+1 +798 p)
1049 is a special instance (+1 -251 p)
1051 is a special instance (-1 +845 p)
1051 is a special instance (-1 -206 p)
1061 is a special instance (-1 +880 p)
1061 is a special instance (-1 -181 p)
1063 is a special instance (+1 +297 p)
1063 is a special instance (+1 -766 p)
1069 is a special instance (-1 +978 p)
1069 is a special instance (-1 -91 p)
1087 is a special instance (+1 +975 p)
1087 is a special instance (+1 -112 p)
1091 is a special instance (-1 +386 p)
1091 is a special instance (-1 -705 p)
1093 is a Wieferich prime
1097 is a special instance (+1 +825 p)
1097 is a special instance (+1 -272 p)

However when I repeat the same with wwwwcl I get
1009 is a special instance (+1 -713 p)
1013 is a special instance (-1 -972 p)
1019 is a special instance (-1 -362 p)
1021 is a special instance (-1 -377 p)
1031 is a special instance (+1 -713 p)
1033 is a special instance (+1 -782 p)
1039 is a special instance (+1 -167 p)
1049 is a special instance (+1 -251 p)
1051 is a special instance (-1 -206 p)
1061 is a special instance (-1 -181 p)
1063 is a special instance (+1 -766 p)
1069 is a special instance (-1 -91 p)
1087 is a special instance (+1 -112 p)
1091 is a special instance (-1 -705 p)
1093 is a Wieferich prime
1097 is a special instance (+1 -272 p)
Now I don't know whether there's something wrong with wwwwcl, but I guess there are indeed two A values for each p and wwwwcl simply omits the larger one. -- Toshio Yamaguchi (tlk−ctb) 19:43, 28 October 2012 (UTC)

There may be two values ≤ 1000 only for primes below 2000. The two absolute values sum up to p. So if p>2000, then at least one of the two is greater than 1000. Maxal (talk) 01:24, 29 October 2012 (UTC)

The last sentence in the section Wieferich prime#Periods of Wieferich primes currently reads
"Garza and Young claim that the period of 1093 were 1092 and that this were the same as the period of 10932,[52]:314 although the fact that the multiplicative order of 2 modulo 10932 is 364 shows that this is not the case."
I am unsure about whether that sentence should be kept or not. I am the editor who originally added that sentence. However, I don't know whether it is appropriate. While I believe the Garza & Young paper is what Wikipedia generally considers a reliable source, I don't know whether that is a reason to repeat that incorrect claim in the article. It seems to be just one particular error in an otherwise reliable source. The note I added to the end of the sentence might come close to WP:OR territory, although I don't know whether a claim such as this one needs a source (I don't know how likely it is  that the statement that the multiplicative order of 2 modulo 10932 is 364 will be challenged). -- Toshio Yamaguchi 11:54, 27 March 2013 (UTC)

I would remove whole section on Periods of Wieferich primes as it repeats what is already said, just in different (recreational math) terms. I am also not sure if Mathematics Magazine is a reliable academic source. For me it sounds like a journal on recreational mathematics, with no so strict reviewing policy. So errors in obvious mathematical facts do not come as a surprise there. Maxal (talk) 01:23, 28 March 2013 (UTC)
I removed the whole section. I agree that Mathematics Magazine doesn't appear to be a reliable academic source. http://www.maa.org/pubs/mm-guide.html says it is an expository journal of undergraduate mathematics and not a research journal. -- Toshio Yamaguchi 06:27, 28 March 2013 (UTC)

I know that for any prime number p, the expression in that section adds up to a number > 0 and < p2 and so is not a multiple of p2. I always thought ≡ a mod m meant it is m * an integer + a. Is that only the definition for when both a and m are integers? In that context, what does the article really mean by ≡ 0 mod p2? There were also so few terms listed in the expression that it's unclear whether it means the reciprocals of all odd numbers up to p-2.  Blackbombchu (talk) 02:28, 3 October 2013 (UTC)

How do you know that the left-hand side of the congruence is always less than p2? That's not really obvious to me. According to Prime number#Euler's analytical proof, the sum of the reciprocals of the primes "becomes bigger than any arbitrary real number provided that p is big enough". That sounds to me as if there exists a break even point somewhere where the sum gets larger than p2. I suppose that the sum is over the reciprocals of all odd prime numbers up to p-2. -- Toshio Yamaguchi 09:26, 3 October 2013 (UTC)
That's right, for any prime number p, that series will eventually pass p but only after you get to the reciprocal of an odd number much higher than p that's even larger than 2p. Blackbombchu (talk) 19:10, 3 October 2013 (UTC)

I think you are right. When p gets large, most of the additional terms on the left-hand side don't really matter anymore. What I think is interesting about this congruence is that obviously it can only hold if the sum of the terms on the left-hand side is an integer. It seems really unlikely to me that this would ever be the case (because the reciprocal of, say, 9 is 0.1111...) and I don't see why there would somehow be a situation where all the digits after the decimal point would add up to 0. -- Toshio Yamaguchi 20:21, 3 October 2013 (UTC)
A different issue in the same section:
"1093 and 3511 are the only primes up to 4 x 1012 satisfying ordp2 2 = ordp 2 and it is known that ord1093 2 = 364 and ord3511 2 = 1755."
In fact, equality ordp2 2 = ordp 2 is equivalent to p being a Wieferich prime (I plan to add this to "Equivalent congruences" section). From this perspective, the quoted statement looks weird. The references should be either moved to "History and search status" if this is about the search or simply removed since what they claim is well known. Maxal (talk) 13:13, 4 October 2013 (UTC)
I agree that this statement is in a sense just saying 1093 and 3511 are the only Wieferich primes up to 4 x 1012, which is already stated elsewhere. My suggestion would be to move the statement about the multiplicative order to Wieferich prime#Explanation of the Wieferich property, as that might not be obvious to someone not being familiar with the subject, so I think it's worth stating it at least somewhere (and that section is intended as a non-technical overview for the layperson, so that statement would be a good fit for that section). -- Toshio Yamaguchi 20:41, 4 October 2013 (UTC)
Explanation of the Wieferich property was an odd section name. I merged it, Equivalent congruences, and parts of Order of 2 modulo powers of Wieferich primes under the section Equivalent definitions. Maxal (talk) 18:33, 5 October 2013 (UTC)

Section 3.3 says 

Therefore, if there are only finitely many Wieferich primes, then there will be at most finitely many Mersenne numbers that are not square-free. Rotkiewicz showed that the converse is also true, that is, if there are infinitely many square-free Mersenne numbers, then there are infinitely many non-Wieferich primes.
One problem is that M6k is always a multiple of 9 so it can not be the case that there are at most finitely many Mersenne numbers that are not square-free. (aside: wouldn't "only" make better sense than "at most"?)
I was about to change it to 

Therefore, if there are only finitely many Wieferich primes, then there will be only finitely many primes q such that the Mersenne number Mq is not square-free.
because that seems to go with what came before and be true.
Then I noticed a second problem: the converse of the current claim would be 

"if there are at most finitely many Mersenne numbers that are not square-free then there are only finitely many Wieferich primes"
equivalently: 

if there are infinitely many Wieferich primes then there are infinitely many Mersenne numbers that are not square-free
But that is not what it says.
So I decided to hold back in hopes that someone could clarify this. With some difficulty I got through to the article but my French is a little rusty. I looked in the review in math reviews which says (after hacking the TeX slightly)

it is proved that (1) if there exist infinitely many square-free Mersenne numbers, then there exist infinitely many primes  
  
    
      
        p
      
    
    {\displaystyle p}
  
 with  
  
    
      
        
          p
          
            2
          
        
      
    
    {\displaystyle p^{2}}
  
 not dividing  
  
    
      
        
          2
          
            p
            −
            1
          
        
        −
        1
      
    
    {\displaystyle 2^{p-1}-1}
  
; (2) there exist infinitely many natural numbers  
  
    
      
        n
      
    
    {\displaystyle n}
  
 with  
  
    
      
        
          n
          
            2
          
        
      
    
    {\displaystyle n^{2}}
  
 dividing  
  
    
      
        
          2
          
            n
          
        
        −
        2
      
    
    {\displaystyle 2^{n}-2}
  
 if and only if there exist infinitely many primes  
  
    
      
        p
      
    
    {\displaystyle p}
  
 with  
  
    
      
        
          p
          
            2
          
        
      
    
    {\displaystyle p^{2}}
  
 dividing 
  
    
      
        
          2
          
            p
            −
            1
          
        
        −
        1
      
    
    {\displaystyle 2^{p-1}-1}
  
.
--Gentlemath (talk) 05:58, 23 November 2013 (UTC)

I know only little French, but theorem 1 on p. 79 in the Rotkiewicz paper appears to state the following in English:
If there exist infinitely many Mersenne numbers without square divisors greater than 1, then there exist infinitely many prime numbers p such that p2 ∤ 2p-1 - 1.
This is also what the article says, though I agree that it is debatable whether that is really the converse of the previously mentioned theorem. -- Toshio Yamaguchi 13:17, 23 November 2013 (UTC)

Maybe this article needs a real shake-up. Also in 3.3 is the claim

For the primes 1093 and 3511 it was shown that neither of them is a factor of any Mersenne or Fermat number.
This is befuddling because, for every prime 
  
    
      
        p
      
    
    {\displaystyle p}
  
 there is a divisor 
  
    
      
        d
      
    
    {\displaystyle d}
  
 of 
  
    
      
        p
        −
        1
      
    
    {\displaystyle p-1}
  
 so that 
  
    
      
        p
      
    
    {\displaystyle p}
  
 divides exactly the Mersenne numbers  
  
    
      
        
          M
          
            n
          
        
      
    
    {\displaystyle M_{n}}
  
 with 
  
    
      
        n
      
    
    {\displaystyle n}
  
 a multiple of 
  
    
      
        d
      
    
    {\displaystyle d}
  
.
In particular the numbers are 
  
    
      
        d
        =
        364
      
    
    {\displaystyle d=364}
  
  for 
  
    
      
        1093
      
    
    {\displaystyle 1093}
  
 and 
  
    
      
        d
        =
        1755
      
    
    {\displaystyle d=1755}
  
  for 
  
    
      
        3511
      
    
    {\displaystyle 3511}
  

I think that the article could be changed to read

For the primes 1093 and 3511 it was shown that neither of them is a factor of any Mersenne number whose index is prime or any Fermat number (at all).
That would make sense with things said a little earlier.
The paper used as a reference is from a good journal and appears to say exactly what is claimed (without my addition). 

THEOREM 2. Neither 1093 nor 3511 divides any 
  
    
      
        
          F
          
            n
          
        
      
    
    {\displaystyle F_{n}}
  
 or any 
  
    
      
        
          M
          
            q
          
        
      
    
    {\displaystyle M_{q}}
  
.
However the paper says at the start 

Throughout, "p" and "q" will denote odd primes.
Still, this problem (and the other) makes me suspicious about the article as a whole.
What a Mersenne number is (perhaps only 
  
    
      
        
          2
          
            n
          
        
        −
        1
      
    
    {\displaystyle 2^{n}-1}
  
 for 
  
    
      
        n
      
    
    {\displaystyle n}
  
 odd? which would make the claim true..) is never explicitly stated. However it follows implicitly from the description of 
  
    
      
        
          2
          
            n
          
        
        −
        1
      
    
    {\displaystyle 2^{n}-1}
  
 as the 
  
    
      
        n
      
    
    {\displaystyle n}
  
th Mersenne number that 
  
    
      
        n
      
    
    {\displaystyle n}
  
 can be even or odd.
--Gentlemath (talk) 06:33, 23 November 2013 (UTC)

The paper by Bray and Warren explicitly talks about Mersenne numbers with prime index, so I adjusted the article to explicitly state that (diff). -- Toshio Yamaguchi 13:28, 23 November 2013 (UTC)

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Please check for and forgive any errors although I've double-checked myself. I do not consider myself a mathematician. The following is merely a suggestion that makes a lot of sense to me but which might be entirely inappropriate for Wikipedia.
The "Equivalent definitions" section shows some detail on the the Fermat quotient with p = 1093 but nowhere in the entire article is the actual Wieferich prime calculation for p = 1093 shown where one divides by p^2 = 1194649.
I assume this was done by hand over a hundred years ago so why not simply do the following somewhere early in the article to show everybody at least one example (out of the two known Wieferich primes) in the most direct and unambiguous way possible of what the article is actually about:
p = 1093
p-1 = 1092
p^2 = 1194649
2^(p-1)-1 =
5305853629099163473739654017028385853919897839577127147455154959874
2698935712572215646062825029533563666321339663466341699370021653261
4451998263607140649559448662636695562122335268130631432841045579576
5830555928325311888973488264278654408606329328273740531145437577728
5526890640894984855797452638426498665888535738081213096656895
(2^(p-1)-1) / (p^2) =
4441349408151819884953366233118167640804870585064840926042004772844
8020243362336732919931147165011282532627859449483774480512704278211
7971051131844701372168267551922527505670983919235383307432597842191
7927823091406188670457589019267294752355151453082654847696216694383
4989934818423641467742786909315203600294760836095968855
Why not include this? It is not particularly long compared to the length of the article and please do not truncate the relatively small long numbers when they only span a handful of lines.
Is the equivalence more important than what it is equivalent to? 90.149.36.98 (talk) 14:41, 10 August 2018 (UTC)

The definition is simple. I think the numbers are too large to be helpful. I have added [5] a small example for another base: 11 is a Wieferich prime base 3 since 311 − 1 − 1 = 59048 = 488 × 112. PrimeHunter (talk) 15:48, 10 August 2018 (UTC)
From the article:
“A Wieferich prime base a is a prime p that satisfies

ap − 1 ≡ 1 (mod p2).
Such a prime cannot divide a, since then it would also divide 1.
It's a conjecture that for every natural number a, there are infinitely many Wieferich primes in base a.”

In the cied article given, I can’t find the term “Wieferich prime base a” although I might have missed it. Furthermore, in all the computations in the cited paper, the number ‘a’ is always less than the prime ‘p’, which is perhaps appropriate, since otherwise, all primes would be ‘Wieferich prime base ‘’ for some a, like ‘a’=p^2-1. So Im going to change the definition to require a<p, but not remove the name or take away citation for now.Rich (talk) 20:48, 25 January 2019 (UTC)

Is there a reason for the "subsequence of" part to be in a serif font? Woah! // Talk? 13:15, 22 December 2022 (UTC)

No. For some reason somebody had used formatting templates intended only for mathematical formulas there. I removed them. —David Eppstein (talk) 07:57, 23 December 2022 (UTC)

The following is a suggested change to the https://en.wikipedia.org/wiki/Wieferich_prime#Wieferich_sequence section.
The following line:
2, 1093, 5, 20771, 18043, 5, 20771, 18043, 5, ..., it gets a cycle: {5, 20771, 18043}.
should be changed to:
2, 1093, 5, 20771, 18043, 5, 20771, 18043, 5, ... (sequence A359952 in the OEIS), it gets a cycle: {5, 20771, 18043}.
Thank you. Boblyonsnj (talk) 17:12, 28 January 2023 (UTC)

 Done small jars tc 11:48, 29 January 2023 (UTC)

In the section on the connections with the abc conjecture, there is the following sentence
"The set of Wieferich primes and the set of non-Wieferich primes, sometimes denoted by W2 and W2c respectively, are complementary sets, so if one of them is shown to be finite, the other one would necessarily have to be infinite, because both are proper subsets of the set of prime numbers."
The final clause seems to not be relevant to the point being made - the fact that they are subsets is (implicitly) implied by complementary, and whether they are proper subsets or not makes no difference as to whether one needs to be finite or not.
Instead the reason is that they are complementary and there are infinitely many primes. I propose replacing the final clause with one to this effect, or if this is deemed not to be at the appropriate level given the surrounding material, simply omitting the clause. 5.151.13.164 (talk) 18:43, 10 April 2024 (UTC)

 Done I removed the "because both" clause. —David Eppstein (talk) 20:36, 10 April 2024 (UTC)

In section History and search status, change the hyperlink for the first ∤ (does not divide) symbol from List of mathematical symbols#notdivide to List of mathematical symbols#∤ in order to fix a broken section link. 94.175.200.195 (talk) 10:35, 27 December 2024 (UTC)

 Done - thank you! ObserveOwl (talk) 16:31, 27 December 2024 (UTC)