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

[Tomasz Ordowski]

Dear Max!

Thank you for finding counterexamples
that force the correction to my conjecture
https://oeis.org/history/view?seq=A366029&v=38

Best,

Thomas

śr., 4 paź 2023 o 18:30 Max Alekseyev <maxale at gmail.com> napisał(a):

> Hi Thomas,
>
> In fact, n = 2^(2^5)+1 is a counterexample to your conjecture.
> It is clearly a Fermat pseudoprime to base 2, and so the congruence modulo
> n^2 holds. It remains to check the congruence modulo 2^n - 1.
>
> Since n-1 = 2^32, we have
> (n-1)^(2^n-2) = 2^(64*(2^(n-1)-1))
> and since 2^n == 1 (mod 2^n - 1), we have
> 2^(32*(2^n-2)) == 2^(64*(2^(n-1)-1) mod n) == 2^0 == 1 (mod 2^n - 1).
>
> The same holds for any other Fermat number.
>
> So you may like to update your conjecture to include any Fermat numbers,
> not just primes.
>
> Regards,
> Max
>
>
> On Tue, Oct 3, 2023 at 11:44 PM Tomasz Ordowski <tomaszordowski at gmail.com>
> wrote:
>
> > 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):
> >
> > > 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.
> > >
> > >
> >
> > --
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> >
>
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