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

[Max Alekseyev]

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/
> >> >
> >>
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >
>
> --
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