## woodall or riesel number theorem

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• Hi, Group, ... Woodall s theorem (as it should be named): ... let R= k*2^n -1, n 1 is a natural number and k
Message 1 of 8 , Sep 13, 2011
Hi, Group,
...
Woodall's theorem (as it should be named):
...
let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
(similar to Proth's theorem.)
...
proof:
if 'm' is from the set of natural numbers, then every odd prime
divisor 'r' of a^(2^m) +1 implies that q == +1(mod a^(m+1))
[concluded from generalized Fermat-number 'proofs' by Proth,
and with me examining Proth's theorem].
...
now, if 'q' is any prime divisor of 'S', then b^((R-1)/2)= (b^k)^
(2^(n-1)) == +1 (mod q) implies that q == +1 (mod 2^n).
...
thus, if 'S' is composite, 'S' will be the product of at least two
primes each of which may have a maximum value of (2^n -1), and it
follows that...
...
k*2^n +1 = (2^n)*(2^n) -1 = (2^n +1)*(2^n -1) <= (2^n)*(2^n) -2*(2^n)
+1; implies that -2 <= -2*(2^n) and dividing by 2^n and multiplying
by -1, we have.. 1 > 2^n, contradiction!
...
hence, for some 'b', if k <= 2^n -1 and b^((R-1)/2) == +1 (mod R),
then 'R' is prime.
*QED
...

Warm regards,

Bill Bouris
the proof can be found at...
www.oddperfectnumbers.com
• oh yeah,... I forgot to mention that b
Message 2 of 8 , Sep 14, 2011
oh yeah,... I forgot to mention that b <= 2*ln(R).
it's further implications... when k=1, means that Mersenne
numbers could be found prime with a single non-pseudo-
prime test (eg. 2^15 mod 31 == 1; you're right; 'b' needs
to be restricted similar to 'k'. Bill (happy hunting!)

From: Bernardo Boncompagni <redgolpe@...>
To: leavemsg1 <leavemsg1@...>
Sent: Wednesday, September 14, 2011 2:08 AM
Subject: Re: [PrimeNumbers] woodall or riesel number theorem

On Wed, Sep 14, 2011 at 8:22 AM, leavemsg1 <leavemsg1@...> wrote:

>let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
>If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
>(similar to Proth's theorem.)
>

n=4, k=11, R=175, b=51,
b^87==1 (mod 175)

[Non-text portions of this message have been removed]
• duh? I thought about it in the car while running some errands; the remedy is that along with gcd(R -1, b)= 1, because gcd(174, 51) = 3. that is the ONLY thing
Message 3 of 8 , Sep 14, 2011
duh? I thought about it in the car while running some
errands; the remedy is that along with gcd(R -1, b)= 1,
because gcd(174, 51) = 3. that is the ONLY thing keep-
ing it from working in your example... enjoy!

--- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...> wrote:
>
> oh yeah,... I forgot to mention that b <= 2*ln(R).
> it's further implications... when k=1, means that Mersenne
> numbers could be found prime with a single non-pseudo-
> prime test (eg. 2^15 mod 31 == 1; you're right; 'b' needs
> to be restricted similar to 'k'. Bill (happy hunting!)
>
> From: Bernardo Boncompagni <redgolpe@...>
> To: leavemsg1 <leavemsg1@...>
> Sent: Wednesday, September 14, 2011 2:08 AM
> Subject: Re: [PrimeNumbers] woodall or riesel number theorem
>
>
>
>
>
> On Wed, Sep 14, 2011 at 8:22 AM, leavemsg1 <leavemsg1@...> wrote:
>
>
> >let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
> >If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
> >(similar to Proth's theorem.)
> >
>
> n=4, k=11, R=175, b=51,
> b^87==1 (mod 175)
>
> [Non-text portions of this message have been removed]
>
• Bernard,   It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it. gcd(174, 51) = 3; otherwise, it holds, not because b
Message 4 of 8 , Sep 14, 2011
Bernard,

It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it.
gcd(174, 51) = 3; otherwise, it holds, not because b <= 2*ln(R). it's a nice
theorem!

Rewards,

Bill

From: Bill Bouris <leavemsg1@...>
To: Bernardo Boncompagni <redgolpe@...>; pgroup <primenumbers@yahoogroups.com>
Sent: Wednesday, September 14, 2011 12:51 PM
Subject: Re: [PrimeNumbers] woodall or riesel number theorem

oh yeah,... I forgot to mention that b <= 2*ln(R).
it's further implications... when k=1, means that Mersenne
numbers could be found prime with a single non-pseudo-
prime test (eg. 2^15 mod 31 == 1; you're right; 'b' needs
to be restricted similar to 'k'. Bill (happy hunting!)

From: Bernardo Boncompagni <redgolpe@...>
To: leavemsg1 <leavemsg1@...>
Sent: Wednesday, September 14, 2011 2:08 AM
Subject: Re: [PrimeNumbers] woodall or riesel number theorem

On Wed, Sep 14, 2011 at 8:22 AM, leavemsg1 <leavemsg1@...> wrote:

>let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
>If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
>(similar to Proth's theorem.)
>

n=4, k=11, R=175, b=51,
b^87==1 (mod 175)

[Non-text portions of this message have been removed]
• It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it. ... n=5, k=21, R=671=11*61, b=9
Message 5 of 8 , Sep 14, 2011
It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it.
> gcd(174, 51) = 3; otherwise, it holds, not because b <= 2*ln(R). it's a
> nice
> theorem!
>

n=5, k=21, R=671=11*61, b=9<2*ln(671), gcd(670,9)=1
b^335==1 (mod 671)

[Non-text portions of this message have been removed]
• only trying to further my theorem; maybe it has to do with gcd(b, 3) = 1, and n 2; just digging; that might need to be the restriction ??? Bill thank you for
Message 6 of 8 , Sep 14, 2011
only trying to further my theorem; maybe it has to do with gcd(b, 3) = 1,
and n >2; just digging; that might need to be the restriction ??? Bill
thank you for investigating it, thus far.

From: Bernardo Boncompagni <redgolpe@...>
Cc: leavemsg1@...
Sent: Wednesday, September 14, 2011 3:23 PM
Subject: Re: [PrimeNumbers] woodall or riesel number theorem

It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it.
>gcd(174, 51) = 3; otherwise, it holds, not because b <= 2*ln(R). it's a nice
>theorem!

n=5, k=21, R=671=11*61, b=9<2*ln(671), gcd(670,9)=1

b^335==1 (mod 671)

[Non-text portions of this message have been removed]
• I only have two strikes against the theorem; 2^2-1 could be the problem, since if R = 3, then 2^((R-1)/2) = 2 mod 3 == -1, and violates the theory. I hope
Message 7 of 8 , Sep 14, 2011
I only have two strikes against the theorem; 2^2-1 could be the problem,
since if R = 3, then 2^((R-1)/2) = 2 mod 3 == -1, and violates the theory.
I hope that's correct--- gcd(b, 3) = 1 should be the hangup! Bill

From: Bill Bouris <leavemsg1@...>
To: Bernardo Boncompagni <redgolpe@...>; pgroup <primenumbers@yahoogroups.com>
Sent: Wednesday, September 14, 2011 4:45 PM
Subject: Re: [PrimeNumbers] Riesel's number theorem

only trying to further my theorem; maybe it has to do with gcd(b, 3) = 1,
and n >2; just digging; that might need to be the restriction ??? Bill
thank you for investigating it, thus far.

From: Bernardo Boncompagni <redgolpe@...>
Cc: leavemsg1@...
Sent: Wednesday, September 14, 2011 3:23 PM
Subject: Re: [PrimeNumbers] woodall or riesel number theorem

It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it.
>gcd(174, 51) = 3; otherwise, it holds, not because b <= 2*ln(R). it's a nice
>theorem!

n=5, k=21, R=671=11*61, b=9<2*ln(671), gcd(670,9)=1

b^335==1 (mod 671)

[Non-text portions of this message have been removed]
• n 2 also...
Message 8 of 8 , Sep 14, 2011
n > 2 also...

--- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...> wrote:
>
> I only have two strikes against the theorem; 2^2-1 could be the problem,
> since if R = 3, then 2^((R-1)/2) = 2 mod 3 == -1, and violates the theory.
> I hope that's correct--- gcd(b, 3) = 1 should be the hangup! Bill
>
> From: Bill Bouris <leavemsg1@...>
> To: Bernardo Boncompagni <redgolpe@...>; pgroup <primenumbers@yahoogroups.com>
> Sent: Wednesday, September 14, 2011 4:45 PM
> Subject: Re: [PrimeNumbers] Riesel's number theorem
>
>
> only trying to further my theorem; maybe it has to do with gcd(b, 3) = 1,
> and n >2; just digging; that might need to be the restriction ??? Bill
> thank you for investigating it, thus far.
>
> From: Bernardo Boncompagni <redgolpe@...>
> Cc: leavemsg1@...
> Sent: Wednesday, September 14, 2011 3:23 PM
> Subject: Re: [PrimeNumbers] woodall or riesel number theorem
>
>
>
>
>
> It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it.
> >gcd(174, 51) = 3; otherwise, it holds, not because b <= 2*ln(R). it's a nice
> >theorem!
> >
>
> n=5, k=21, R=671=11*61, b=9<2*ln(671), gcd(670,9)=1
>
> b^335==1 (mod 671)
>
> [Non-text portions of this message have been removed]
>
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