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

[Tomasz Ordowski]

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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>>
>