[seqfan] Re: Numbers n such that 2^(2^n-2) == 1 (mod n^2)

[Max Alekseyev]

PS. Wieferich prime p=1093 cannot divide such n, since it would require
ord_{p^2}(2) = 364 = 2^2 * 91 to divide 2^n - 2, which is impossible.


On Wed, Jul 24, 2024 at 3:15 PM Max Alekseyev <maxale at gmail.com> wrote:

> Let n be an integer such that 2^(2^n-2) == 1 (mod n^2) and 2^(n-1) =/= 1
> (mod n). Clearly, n is odd.
> Consider any prime p|n, and let d:=v_p(n).
> Then 2^(2^n-2) == 1 (mod p^{2d}). If p is not Wieferich prime, then
> p^{2d-1} | 2^n-2 = 2*(2^(n-1)-1), and thus p^d | 2^(n-1)-1.
> If all primes p|n are not Wieferich, it would follow that n | 2^(n-1)-1, a
> contradiction.
> Hence, n must have a Wieferich prime factor.
>
> Regards,
> Max
>
>
> On Wed, Jul 24, 2024 at 11:53 AM Tomasz Ordowski <tomaszordowski at gmail.com>
> wrote:
>
>> PS. Numbers n such that 2^(2^n-2) == 1 (mod n^2) and 2^(n-1) =/= 1 (mod
>> n):
>> 66709, 951481, 2215441, 2847421, 4111381, 4869757, 28758601, 81844921,
>> 124187581, 300510001, 306197821, 1221936841, ...  [Amiram Eldar].
>> All these numbers are multiples of the Wieferich prime 3511.
>> How to prove that this must be the case in general?
>>
>> Tom Ordo
>>
>> pt., 19 lip 2024 o 18:05 Tomasz Ordowski <tomaszordowski at gmail.com>
>> napisał(a):
>>
>> > Emmanuel, thanks for this next exception.
>> >
>> > Note that A069051 > 2 is a subsequence.
>> > https://oeis.org/history/view?seq=A069051&v=116
>> > See my new comment.
>> >
>> > Also A217468 is a subsequence.
>> > https://oeis.org/A217468
>> >
>> > Tom Ordo
>> >
>> >
>> > pt., 19 lip 2024 o 11:57 Emmanuel Vantieghem <
>> emmanuelvantieghem at gmail.com>
>> > napisał(a):
>> >
>> >> 951481 = 271*3511  is the next exception.
>> >>
>> >> Op do 18 jul 2024 om 23:57 schreef Tomasz Ordowski <
>> >> tomaszordowski at gmail.com
>> >> >:
>> >>
>> >> > 1, 3, 7, 19, 43, 73, 127, 163, 337, 341,
>> >> > 379, 487, 601, 881, 883, 937, 1387, ...
>> >> > These numbers are not in the OEIS.
>> >> > Such composites 341, 1387, 4681,
>> >> > 5461, 8911, 10261, 14491, 15841, ...
>> >> > are Fermat pseudoprimes to base 2,
>> >> > except the number 66709 = 19*3511,
>> >> > that Amiram Eldar found. Note that
>> >> > the factor 3511 is a Wieferich prime.
>> >> >   Are there any other exceptions?
>> >> >
>> >> > If p is prime and 2^(2^p-2) == 1 (mod p),
>> >> > then 2^(2^p-2) == 1 (mod p^2).
>> >> >
>> >> >    Generally,
>> >> > if 2^(2^n-2) == 1 (mod n),
>> >> > then 2^(2^n-2) == 1 (mod n^2)
>> >> > if and only if 2^(n-1) == 1 (mod n),
>> >> > with the exception n = 66709 = 19*3511.
>> >> >    Find more counterexamples.
>> >> >
>> >> > Best,
>> >> >
>> >> > Tom Ordo
>> >> >
>> >> > --
>> >> > Seqfan Mailing list - http://list.seqfan.eu/
>> >> >
>> >>
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>> >>
>> >
>>
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