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

[Tomasz Ordowski]

Dear readers!

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.