From: Tim Little on
On 2010-07-24, Sebastian Garth <sebastiangarth(a)gmail.com> wrote:
> If you're asserting that, generally speaking, ((Q mod s(N, 1)) mod N)
> = (Q mod N), always, for all Q, then that's just plain wrong.

I'm not. It holds for all odd N (as s(N,1) is divisible by N), and
the even case is irrelevant as the equality never holds. The
simplified equation is true iff the original one is.


- Tim
From: Sebastian Garth on
On Jul 24, 2:07 am, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-07-24, Sebastian Garth <sebastianga...(a)gmail.com> wrote:
>
> > If you're asserting that, generally speaking, ((Q mod s(N, 1)) mod N)
> > = (Q mod N), always, for all Q, then that's just plain wrong.
>
> I'm not.  It holds for all odd N (as s(N,1) is divisible by N), and
> the even case is irrelevant as the equality never holds.  The
> simplified equation is true iff the original one is.
>
> - Tim

Of course! Yes, originally my equation accounted for all N > 2 (for
the sake of "completeness"). Thanks for the clarification.

Cheers,

- Sebastian
From: Frederick Williams on
Sebastian Garth wrote:
>
> In an earlier thread, I put forward a conjecture that generalized
> Fermat's Little Theorem. Specifically:
>
> For all N > 2, IFF gcd(s(N, N - 1) mod s(N, 1), N) = 1 then N is
> either a prime or a Carmichael number, where s(N, E) is the sum of
> powers (eg: 1^E + 2^E ... + N^E).

What does

Iff ... then ...

mean? I am familiar with

If ... then ... and ... iff ...

--
I can't go on, I'll go on.
From: spudnik on
if & only if, that is to say,
Liebniz's neccesity & sufficiency,
to be used in any literate manner!

>   Iff ... then ...

--les ducs d'oil!
http://tarpley.net
From: Gerry Myerson on
In article
<050907ee-6b56-4100-8423-08c2f3d419d9(a)d37g2000yqm.googlegroups.com>,
Sebastian Garth <sebastiangarth(a)gmail.com> wrote:

> In an earlier thread, I put forward a conjecture that generalized
> Fermat's Little Theorem. Specifically:
>
> For all N > 2, IFF gcd(s(N, N - 1) mod s(N, 1), N) = 1 then N is
> either a prime or a Carmichael number, where s(N, E) is the sum of
> powers (eg: 1^E + 2^E ... + N^E).
>
> Using a related concept, I can now make a statement that generalizes
> *all* prime numbers:
>
> For all N > 2, IFF ((s(N, N - 1) mod s(N, 1)) + 1) mod N = 0 then N is
> definitely prime, where s(N, E) is the sum of powers (eg: 1^E +
> 2^E ... + N^E).
>
> AFAIK, the only other theorem that achieves a similar level of
> concision is Wilson's Theorem, so the implications of this equation
> may be quite significant (eg: may lead to much better primality
> tests).

You may be interested in looking up Giuga's conjecture.

--
Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)