Remarks about Wieferich primes in relation with my proof about squarefreeness of Fermat and Mersenne numbers. (Part 1)
 
 

Jacques Basaldúa, 2 January 2002


    Since I released my proof (19/12/2001), there have been many reactions to it. I warmly thank those who have contributed to this. In the future I will sort and publish contributions and references to my work. This, the first part of my answer to the most important contribution, is just a "draft" to avoid keeping inaccurate material in my website. It will take some time to find (I hope), a way to integrate Wieferich primes in my scheme, so I can't yet compromise a date for part 2. At least, this can be said already:
  
    Jack Brennen posted the following note at yahoo groups ( http://groups.yahoo.com/group/primenumbers/message/4609 ):
 
    Yes, there is a flaw. I can't pinpoint the exact error in reasoning (basically, I don't want to spend the time to find it :-), but the equation which is "proven" in section 4.2 is not true. There is an easily computable counterexample.

    The author states that:

    ß(p^n) = p^(n-1)·ß(p) (for any prime p)

    This is not true if p = 1093 & n = 2:
    ß(p^n) == 364
    p^(n-1)·ß(p) == 1093*364

    I believe that in general, the postulate in section 4.2 is not true for any Wieferich prime.

    Somebody apparently tried to prove the Mersennes squarefree using this technique in '96 and came up against this same problem:
http://www2.netdoor.com/~acurry/mersenne/archive2/0037.html
 
    The note contains two important and new (for me) issues:

    1. Precedents: There is some precedent of my proof, but to my knowledge, only about the question of the squarefreeness of Mersenne numbers.

     2. Wieferich primes: What happens with n = 10932 and 35112 ? (Part 1)

1. Precedent
  
    Will Edgington in September 96 stated that if no prime n satisfied 2n-1 - 1 º 0 (mod n2) , this would prove the squarefreeness of Mersenne numbers. When he found out that there were only two known exceptions: the primes 1093 and 3511, he showed that those numbers could not be factors of Mersenne numbers. (This is even easier in my scheme, where ß(Mp) = p must be prime and ß(1093) = 384, ß(3511) = 1755.)

    Of course, since no other Wieferich prime is known, and nobody knows how to find one, we can say that "There is no margin for Mersenne numbers to be anything else but squarefree." But unless we find the way to integrate Wieferich primes in an analyzable scheme, the proof is, as Jack Brennen claims, flawed.

    So the proof does belong to Mr. Edgington and to the person who proves that any number n satisfying 2n-1 - 1 º 0 (mod n2) (not just the two known cases) cannot be a factor of a Mersenne number. Nevertheless, there is much more than that in my paper as I point out just below.
 

2. Wieferich primes
 
    2.1 Mea culpa

    The first impression one may get after reading Mr. Brennen's note is that I did not verify my formula. The formula (and the other hypotheses) were verified for all numbers £ 1 000 000 and all numbers whose ß £ 64 (The biggest one is 18 446 744 073 709 551 615 ). I have a precomputed database of ß and primality results for those numbers which makes any verification very fast. 10932 is 1 194 649 , which is not much bigger, but is above that range. As far as we know, there are only two exceptions to the formula ß(n) = LCM(..) so the consequences of the formula are still valid, although the proofs do require the "Wieferich exception" to be solved. Mr. Brennen did not find an exception, he knew one, but the general rule is: numbers do satisfy the formula.

    As it is often the case, the formula was obtained from the results (and the fact that ß must divide Phi), but there still is a proof, so where is the flaw?

    Trying to follow the proof with numbers, we will get the flaw:

    The units mod 1093 form three 364 unit long subgroups:
 

Units 2i mod 1093   0 : 1, 2, 4, 8, 16, 32, 64, 128, 256, 512,  
 10 : 1024, 955, 817, 541, 1082, 1071, 1049, 1005, 917, 741,
 20 : 389, 778, 463, 926, 759, 425, 850, 607, 121, 242,
 30 : 484,  .  .  .
 250 : 95, 190, 380, 760, 427, 854, 615, 137, 274, 548,
260 : 3 , 6, 12, 24, 48, 96, 192, 384, 768, 443,
270 : 886, 679, 265, 530, 1060, 1027, 961, 829, 565, 37,
280 : 74,  .  .  .
350 : 298, 596, 99, 198, 396, 792, 491, 982, 871, 649,
360 : 205, 410, 820, 547
Units 5·2i mod 1093   0 : 5, 10, 20, 40, 80, 160, 320, 640, 187, 374,
 10 : 748, 403, 806, 519, 1038, 983, 873, 653, 213, 426,
 20 : 852,  .  .  .
 350 : 397, 794, 495, 990, 887, 681, 269, 538, 1076, 1059,
360 : 1025, 957, 821, 549
Units 7·2i mod 1093   0 : 7, 14, 28, 56, 112, 224, 448, 896, 699, 305,
 10 : 610, 127, 254, 508, 1016, 939, 785, 477, 954, 815,
 20 : 537,  .  .  .
 350 : 993, 893, 693, 293, 586, 79, 158, 316, 632, 171,
360 : 342, 684, 275, 550

.
     The units mod 10932 form 1092 subgroups of length 364 from which the first two are:
 

Units 2i mod 10932   0 : 1, 2, 4, 8, 16, 32, 64, 128, 256, 512,
 10 : 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288,
 20 : 1048576, 902503, 610357, 26065, 52130, 104260, 208520, 417040, 834080, 473511,
 30 : 947022,  .  .  .
350 : 118342, 236684, 473368, 946736, 698823, 202997, 405994, 811988, 429327, 858654,
360 : 522659, 1045318, 895987, 597325
Units 3·2i mod 10932   0 : 3 , 6, 12, 24, 48, 96, 192, 384, 768, 1536,
 10 : 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 378215,
 20 : 756430, 318211, 636422, 78195, 156390, 312780, 625560, 56471, 112942, 225884,
 30 : 451768,  .  .  .
350 : 355026, 710052, 225455, 450910, 901820, 608991, 23333, 46666, 93332, 186664,
360 : 373328, 746656, 298663, 597326
        ...   .  .  .

.
     Notice the unit 3. It is red modulo 1093 and it is green modulo 10932. We pretended to prove that, if it belonged to one subgroup modulo p, it had to belong to the same subgroup modulo pn.

    This is:

    3 = 2i mod 1093 = 2260 mod 1093  (note the value 260 is in the column containing exponents.)

    We pretended to prove that there existed some j satisfying:

    3 = 2j mod 10932

    and we proceeded as follows:

    3 = 2260 - (2260 div 1093)·1093 = 2j - (2j div 10932)·10932 = 2j - (2j div 10932)·1093·1093
    2j = 2260 - (so - s2)·1093
    where so - s2 is (2260 div 1093) - (2j div 10932)·1093

    The difference between 2j and 2260 is something times 1093. This implies the congruence modulo 1093, but it does not imply the existence of such a j. And in this case, there is no j such that 3 = 2j mod 10932.

    The proof is not valid, but the formula does work (almost always).
  
   2.2 And now .. The good news!

    I discovered the conjecture about the squarefreeness of Fermat and Mersenne numbers quite accidentally. (After I wrote my article I have found even more references including, of course, Mr. Edgington's.) I believed to have a formal proof for it and decided to publish it. Nevertheless, my interest in the subject is much more directed to the relation between the numbers considered as shapes (which I informally refer to as logic relations) and the numbers considered as numbers (which I informally refer to as arithmetic relations). The confrontation of a logic relation such as ß(Fk) = 2k+1 (and all the others in chap 3) with an arithmetic relation like ß(n) = LCM(ai-1·ß(a),  ... ), the fact that both must simultaneously meet, can be used to prove many number theory issues and surely will.

    If someone still doubts it after reading chapter 3, note the following binary representations:
 
        Phi(1093) = 1092 = 10001000100
        Phi(3511) = 3510 = 110110110110
    The only two known exceptions to the arithmetic relation are the repetition of the patterns 1000 and 110. It looks as if one could find an exception to the general arithmetic rule by choosing the "appropriate patterns". This makes the problem more interesting than it has ever been, suggesting there are still "simple" things which are not understood and offering highly exciting fields of research.

    And, if that wasn't enough, there is more:

    A. Wieferich proved in "Zum letzten Fermat'schen Theorem", 1909, that if Fermat's equation xp + yp + zp = 0 has an integer solution, with x, y, z non zero, for some exponent p, then 2p-1 = 1 mod p2. This suggests further knowledge about Wieferich primes could help finding a comprehensive proof of Fermat's Last Theorem.
 
    Madrid, 2 January 2002.