# ++Eureka!!!

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• 07-23-2010
Sebastiani
++Eureka!!!
Well folks, I've really done it now!

In an earlier thread, I put forward a conjecture that generalized Fermat's Little Theorem. Using a related concept, I can now make a statement that generalizes *all* prime numbers!

Here it is:

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).
• 07-25-2010
Sebastiani
It turns out that one of those terms is completely unnecessary. A simplified description of both conjectures:

Define g(N) = 1^N-1 + 2^N-1 ... + N^N-1.

Conjecture #1:

Iff (g(N) + 1) mod N = 0, then N is prime.

Conjecture #2:

Iff gcd(g(N), N) = 1, then N is either prime or an absolute Fermat pseudoprime (Carmichael number).
• 07-25-2010
pianorain
Even though I haven't posted on either thread, I find this very interesting and would like to hear more about it. Are you close to having a proof of any of your conjectures?
• 07-25-2010
Sebastiani
Quote:

Originally Posted by pianorain
Even though I haven't posted on either thread, I find this very interesting and would like to hear more about it. Are you close to having a proof of any of your conjectures?

Unfortunately, no - producing mathematical proofs is not a strength of mine. But I am working on it.
• 07-25-2010
phantomotap
Quote:

Define g(N) = 1^N-1 + 2^N-1 ... + N^N-1.
The "N^N-1" term is never useful by definition and the most expensive in practice; get rid of it.

(And if you actually write code to test it, don't forget to eliminate the majority using the trivial (+1/-1) mod 6 cases.)

O_o

I wonder if the trivially "definitely not prime" test can be used on the other terms as well?!
[/Edit]

Quote:

Are you close to having a proof of any of your conjectures?
I think this qualifies as a generalization of a test by summation. If so, the proof is exactly the same as Fermat's Little Theorem. (He is basically running every relevant "Fermat Primality Test" at the "same time".) Then again, I have a terrible headache and may have completely misunderstood what he is doing. ;_;

Soma
• 07-25-2010
Sebastiani
Quote:

Originally Posted by phantomotap
The "N^N-1" term is never useful by definition and the most expensive in practice; get rid of it.

Yep, you're right. Good point.

Quote:

Originally Posted by phantomotap

(And if you actually write code to test it, don't forget to eliminate the majority using the trivial (+1/-1) mod 6 cases.)

Another good point.

Quote:

Originally Posted by phantomotap

I wonder if the trivially "definitely not prime" test can be used on the other terms as well?!

Could you elaborate on that?

Quote:

Originally Posted by phantomotap
I think this qualifies as a generalization of a test by summation. If so, the proof is exactly the same as Fermat's Little Theorem. (He is basically running every relevant "Fermat Primality Test" at the "same time".) Then again, I have a terrible headache and may have completely misunderstood what he is doing. ;_;

Hmm...
• 07-25-2010
phantomotap
Quote:

Could you elaborate on that?
*shrug*

Not really; it was just a little "bing" that... "binged" while I wrote my post.

I'm basically wondering if some of the other factors couldn't be "omitted" because they either contribute nothing or result in a term that naturally requires the sum to not be "mod P".

Soma
• 07-26-2010
EVOEx
Quote:

Originally Posted by Sebastiani
It turns out that one of those terms is completely unnecessary. A simplified description of both conjectures:

Define g(N) = 1^N-1 + 2^N-1 ... + N^N-1.

Conjecture #1:

Iff (g(N) + 1) mod N = 0, then N is prime.

Conjecture #2:

Iff gcd(g(N), N) = 1, then N is either prime or an absolute Fermat pseudoprime (Carmichael number).

I can prove it one way... Euler's theorem - Wikipedia, the free encyclopedia
a^phi(n) = 1 (mod n) iff a and n are relative prime. So if p is prime, since phi(p) = p - 1:
a^(p-1) = 1 (mod n)
So g(n), when n is prime, is 1 + 1 + ... + 1 (mod n) = n-1 (mod n).

The other way around... Not sure if it's really true, to be honest. I doubt it ;-).

And it's of course very unlikely for g(N) = 1 (mod N) for any other number (1 in N) if it's distributed equally, so it's quite difficult to find a counterexample probably as well. But I think there is a counter example.

Edit: Note that g(N) (mod N) = 2*(1^(N-1) + 2^(N-1) + ... + ((N-1)/2)^(N-1)) (mod N)
• 07-27-2010
Epy
• 07-27-2010
pianorain
Quote:

Originally Posted by Epy

This is an odd thing to say considering that Paul Erdős was born in 1913.
• 07-27-2010
EVOEx
Quote:

Originally Posted by pianorain
This is an odd thing to say considering that Paul Erdős was born in 1913.

He was just very talented... ;)

Really, it's a load of crap, though. There are infinitely many unknown things in mathematics, even important thing that we don't know or can be improved. And once we know more, there is more that we need as it only begs more questions.
For instance, take the poincare conjecture. It's been solved only fairly recently. And there are MANY more unsolved questions. For a small selection: Unsolved Problems -- from Wolfram MathWorld
• 07-27-2010
DavidP
Quote:

Originally Posted by Epy

On the contrary, if you're really serious about this, why don't you start writing an academic paper, get it published, and have a mathematics professor either co-author it or just at least be nice enough to look over it for you to make sure it's fool proof?
• 07-28-2010
Epy
Quote:

Originally Posted by DavidP
On the contrary, if you're really serious about this, why don't you start writing an academic paper, get it published, and have a mathematics professor either co-author it or just at least be nice enough to look over it for you to make sure it's fool proof?

That statement doesn't really make any sense, you're forming it as if you're disagreeing with me but you're really not. He should write a paper, not post every detail to a forum. He's counting on all of us being slackers.
• 07-28-2010
Epy
Quote:

Originally Posted by pianorain
This is an odd thing to say considering that Paul Erdős was born in 1913.

Never read anything he's done. Everything I've ever looked at has been by Euler, Poincare, Gauss, Ramanujan, Fermat, etc. The foundations of mathematics were certainly done before the 1900s, in my opinion. (Yes, I know that Ramanunjan was active into the early 1900s, but he was all number theory stuff anyway)
• 07-28-2010
laserlight
Quote:

Originally Posted by Epy