## Prime Conjecture

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• I have found this new prime conjecture n is a prime number only when k is a hole number ((2^n)-2)/n = k Discovered by: Eduardo Mourey López Negrete Torreón
Message 1 of 7 , Jun 9, 2004
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I have found this new prime conjecture

n is a prime number only when k is a hole number

((2^n)-2)/n = k

Discovered by:
Eduardo Mourey López Negrete
Torreón Coah. Mexico
Torreón Jardin
Lirios 333
27200
• If n is a prime number then ((2^n)-2)/n is always an integer. This follows from Fermats Little Theorem, one of the first things one learns about in number
Message 2 of 7 , Jun 9, 2004
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If n is a prime number then ((2^n)-2)/n is always an integer. This
follows from Fermats Little Theorem, one of the first things one learns
about in number theory. The converse is false, ie the statement that:

If ((2^n)-2)/n is an integer then n is prime

is not right. For instance, consider n=341 then you get

(2^341 - 2)/341 =
1313633279869679888889995472
4741608669335164206654835981
8181178942157881007634073042
86671514789484550 (one big integer spanning over 4 rows)

which clearly is an integer. However, 341 is not prime.
http://mathworld.wolfram.com/FermatsLittleTheorem.html

Regards,
/m

edmorrey wrote:
> I have found this new prime conjecture
>
> n is a prime number only when k is a hole number
>
> ((2^n)-2)/n = k
>
>
>
> Discovered by:
> Eduardo Mourey López Negrete
> Torreón Coah. Mexico
> Torreón Jardin
> Lirios 333
> 27200
>
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
> <http://rd.yahoo.com/SIG=129m9q24q/M=298184.5022502.6152625.3001176/D=groups/S=1705083388:HM/EXP=1086885375/A=2164331/R=0/SIG=11eaelai9/*http://www.netflix.com/Default?mqso=60183351>
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• ... Hi, this is the fermat-test for base 2: n is prime only if 2^(n-1) = 1 mod (n) == (2^(n-1)-1)/n is an integer == 2*(2^(n-1)-1)/n=(2^n-2)/n is an integer.
Message 3 of 7 , Jun 9, 2004
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> I have found this new prime conjecture
>
> n is a prime number only when k is a hole number
>
> ((2^n)-2)/n = k

Hi,

this is the fermat-test for base 2:

n is prime only if 2^(n-1) = 1 mod (n) ==> (2^(n-1)-1)/n is an integer
==> 2*(2^(n-1)-1)/n=(2^n-2)/n is an integer.

Cyrix
• I believe Fermat discovered this centuries ago (or possibly even Euclid?). This is simply a Fermat-2 PRP test. Yes, all primes certainly ARE shown by k
Message 4 of 7 , Jun 9, 2004
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I believe Fermat "discovered" this centuries ago (or possibly even
Euclid?). This is simply a Fermat-2 PRP test. Yes, all primes
certainly ARE shown by k being a whole number (i.e. ((2^n)-2)%n == 0)

This can be extened for any "base". So ((2^n)-2)%n==0 implies
Fermat PRP in base 2, and ((b^n)-b)%n==0 implies Fermat PRP in base
b. NOTE these are still just "probably" prime. For your PRP-2
below, 341 is the smallest composite that "passes". However, as
you pointed out, there will be NO primes that fail this test.

This method is most "commonly" used in modular space. Computing
whether ((2^329018309280127)-2)/329018309280127 is a whole number
(without using modular methods), is a VERY hard thing to do, since
2^329018309280127 is a number larger than any "current" computer can
handle. However, computing (2^329018309280127)%329018309280127
then subtracting 2, and checking if the result is zero, can be done
on binary computers in a fraction of a second, by continually
reducing the intermediate results mod n (or 329018309280127 in this
case).

It is always fun to "discover" something, however, when I have done
so in the past, I usually (always in my case) find out that someone,
(like Euclid, Fermat, Erdos, Euler, Newton, Pascal, ...) beat me to
it long long ago. However, I still keep trying ;)

Jim.

--- In primenumbers@yahoogroups.com, "edmorrey" <edmorrey@y...> wrote:
> I have found this new prime conjecture
>
> n is a prime number only when k is a hole number
>
> ((2^n)-2)/n = k
>
>
>
> Discovered by:
> Eduardo Mourey López Negrete
> Torreón Coah. Mexico
> Torreón Jardin
> Lirios 333
> 27200
• Hello All: Can anyone prove this conjecture? For all n 2 ; for all i: 1
Message 5 of 7 , Mar 2, 2008
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Hello All:

Can anyone prove this conjecture?

For all n>2 ;

for all i: 1<= i<= n exist j: n+1 <= j <= n^(3/2)

(1+pj)/(-1+pi) is integer.

pi is the ith prime number.

Sincerely:

Sebastian Martín Ruiz

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• ... Yes, yes. Maybe I ll share that proof later, hehe. Along similar lines: Given a prime p, is there always an r
Message 6 of 7 , Mar 4, 2008
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--- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
wrote:
>
> Hello All:
>
> Can anyone prove this conjecture?
>
> For all n>2 ;
>
> for all i: 1<= i<= n exist j: n+1 <= j <= n^(3/2)
>
> (1+pj)/(-1+pi) is integer.
>
>
> pi is the ith prime number.
>
> Sincerely:
>
> Sebastian Martín Ruiz

Yes, yes. Maybe I'll share that proof later, hehe.

Along similar lines:
Given a prime p, is there always an r < (log(2*p+1))^2
such that p*r - 1 is prime?
(Or such that p*r + 1 is prime?)

Example: let p = 740191.
The first r such that p*r - 1 is prime is r = 192.
But (log(2*740191+1))^2 =~201.9 > 192.

Mark
• Sebastian asked the following. 1a. Prime Conjecture Posted by: Sebastian Martin sebi_sebi@yahoo.com sebi_sebi Date: Sun Mar 2, 2008 8:25 am ((PST)) Hello
Message 7 of 7 , Mar 4, 2008
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1a. Prime Conjecture
Posted by: "Sebastian Martin" sebi_sebi@... sebi_sebi
Date: Sun Mar 2, 2008 8:25 am ((PST))

Hello All:

Can anyone prove this conjecture?

For all n>2 ;

for all i: 1<= i<= n exist j: n+1 <= j <= n^(3/2)

(1+pj)/(-1+pi) is integer.

pi is the ith prime number.

Sincerely:

Sebastian Martín Ruiz

Sebastian, this should be quite easy to prove for just about any subsequence of the positive integers.

Prove that it's true for the entire set of positive integers, then it would be trivial to show that it's
also true for the prime number subsequence.

Kermit Rose < kermit@... >
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