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

[Emmanuel Vantieghem]

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
>
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