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

Tomasz Ordowski tomaszordowski at gmail.com
Sun Oct 1 20:18:53 CEST 2023

```PS. My conjecture (more precisely):

For n > 1, (2^n-1)n^2 divides (n-1)^(2^n-2)-1 if and only if
2^n-1 is a (Mersenne) prime or n = 2^(2^k)+1 is a (Fermat) prime.

It is necessary to check whether n = 2^(2^5)+1 is not a counterexample.

Best,

Thomas
______________________
https://oeis.org/history/view?seq=A366029&v=26

niedz., 1 paź 2023 o 17:21 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

>
> Conjecture:
> For n > 1, (2^n-1)n^2 divides (n-1)^(2^n-2) - 1
> if and only if 2^n - 1 is prime or n = 257 and 65537.
>
> Is n = 65537 the second and probably final exception?
> As the largest known Fermat primes F_4 = 2^2^4 + 1.
>
> See my draft A366029:
> https://oeis.org/draft/A366029
> https://oeis.org/history/view?seq=A366029&v=24
>
> Best regards,
>
> Thomas Ordowski
> ___________
> Amiram Eldar:
> Yes, 65537 is also a term.
> I tried to check 2^32+1 but it seems that is too high to check.
>
>
```