|
|
 Up |
|
|
  |
Author: JSHJSH Date: Jun 8, 2008 19:22
After my latest miss-steps I went back, looked everything over again
and finally noticed something that was bizarrely, simply obvious and
staring me in the face.
Here's the factoring algorithm that results.
Given an odd target composite T, for a test odd prime p, iterate
through non-zero integers k until
T - k^2 mod p
is a quadratic residue. If no k will work, discard that prime p and
try another. If one does work then get α from
α^2 = (T - k^2)(k^2)^{-1} mod p.
Then you have residues of your factors modulo p from
f_1 = αk mod p
and
f_2 = α^{-1}(1 + α^2)k mod p
where f_1*f_2 = T.
Trivial factorizations if they are given can be discarded by noting
that your residues is the same as T, though one caveat is that for a
given prime p, one of your factors may equal the residue of the target
composite T.
|
| Show full article (1.76Kb) |
|
| | 15 Comments |
|
|
Author: Ivar RosquistIvar Rosquist Date: Jun 8, 2008 19:46
On Sun, 08 Jun 2008 19:22:54 -0700, JSH wrote:
> For example, if 80 primes are found then a target at least as large as
> 80! would topple even if you took primes starting at 3 and working your
> way up, so assuming about 50%% would work then using only primes under
> 1000, you could factor a number greater than
> 7.1569457046263802294811533723187e+118.
OK, for the nth time, factorize
5E7153F9EA8C67E08660CE08001D42715D6008A1718CDAD2F67D79844FD0013ACBDD05
081A1A26AF8AD662EFF37E47FB1FB33D8F7E66207FEEB439B2665EBD87745DC4BF4332
038A55E5F3C110E2CA6CB96689C2EF15C40C40AE6D2F3128B2423A11BFAB030F8D1386
9BF4B4AB367D26DDB90A47FC0ECCEB2BA7D0C6EC21B7A7
I bet your pathetic method can't tackle this before the end of
the universe. Come on, you nitwit, prove me wrong.
|
| |
|
| | no comments |
|
|
Author: Gib BogleGib Bogle Date: Jun 9, 2008 00:16
JSH wrote:
> After my latest miss-steps I went back, looked everything over again
> and finally noticed something that was bizarrely, simply obvious and
> staring me in the face.
Interesting concept of what's bizarre.
|
| |
| no comments |
|
|
Author: Gib BogleGib Bogle Date: Jun 9, 2008 00:17
Ivar Rosquist wrote:
> On Sun, 08 Jun 2008 19:22:54 -0700, JSH wrote:
>
>> For example, if 80 primes are found then a target at least as large as
>> 80! would topple even if you took primes starting at 3 and working your
>> way up, so assuming about 50%% would work then using only primes under
>> 1000, you could factor a number greater than
>> 7.1569457046263802294811533723187e+118.
>
> OK, for the nth time, factorize
>
> 5E7153F9EA8C67E08660CE08001D42715D6008A1718CDAD2F67D79844FD0013ACBDD05
> 081A1A26AF8AD662EFF37E47FB1FB33D8F7E66207FEEB439B2665EBD87745DC4BF4332
> 038A55E5F3C110E2CA6CB96689C2EF15C40C40AE6D2F3128B2423A11BFAB030F8D1386
> 9BF4B4AB367D26DDB90A47FC0ECCEB2BA7D0C6EC21B7A7
>
> I bet your pathetic method can't tackle this before the end of
> the universe. Come on, you nitwit...
|
| Show full article (0.87Kb) |
| no comments |
|
|
Author: rossumrossum Date: Jun 9, 2008 03:57
On Sun, 8 Jun 2008 19:22:54 -0700 (PDT), JSH gmail.com> wrote:
>After my latest miss-steps I went back, looked everything over again
>and finally noticed something that was bizarrely, simply obvious and
>staring me in the face.
>
>Here's the factoring algorithm that results.
>
>Given an odd target composite T, for a test odd prime p,
There are an infinite number of odd primes p. What limits would you
suggest for the search: 3 <= p <= ???. Searching an infinite number
of primes can take an infinite amount of time.
|
| Show full article (3.00Kb) |
| no comments |
|
|
Author: EnricoEnrico Date: Jun 9, 2008 07:09
On Jun 8, 8:22�pm, JSH gmail.com> wrote:> After my latest miss-steps I went back, looked everything over again
> and finally noticed something that was bizarrely, simply obvious and
> staring me in the face.
>
> Here's the factoring algorithm that results.
>
> Given an odd target composite T, for a test odd prime p, iterate
> through non-zero integers k until
>
> T - k^2 mod p
>
> is a quadratic residue. If no k will work, discard that prime p and
> try another. �If one does work then get � from
>
> �2 = (T - k^2)(k^2)^{-1} mod p.
>
> Then you have residues of your factors modulo p from
>
> f_1 = �k mod p ...
|
| Show full article (2.09Kb) |
| no comments |
|
|
Author: JSHJSH Date: Jun 9, 2008 07:11
On Jun 9, 3:57 am, rossum coldmail.com> wrote:> On Sun, 8 Jun 2008 19:22:54 -0700 (PDT), JSH gmail.com> wrote:>>After my latest miss-steps I went back, looked everything over again
>>and finally noticed something that was bizarrely, simply obvious and
>>staring me in the face.
>
>>Here's the factoring algorithm that results.
>
>>Given an odd target composite T, for a test odd prime p,
>
> There are an infinite number of odd primes p. What limits would you
> suggest for the search: 3 <= p <= ???. Searching an infinite number
> of primes can take an infinite amount of time.
You start with p=3 and work up from there until the product of your
primes that work is greater than the square root of the target.
|
| Show full article (2.84Kb) |
| no comments |
|
|
Author: EnricoEnrico Date: Jun 9, 2008 07:54
On Jun 9, 8:11 am, JSH gmail.com> wrote:> On Jun 9, 3:57 am, rossum coldmail.com> wrote:
>>> On Sun, 8 Jun 2008 19:22:54 -0700 (PDT), JSH gmail.com> wrote:>>>After my latest miss-steps I went back, looked everything over again
>>>and finally noticed something that was bizarrely, simply obvious and
>>>staring me in the face.
>
>>>Here's the factoring algorithm that results.
>
>>>Given an odd target composite T, for a test odd prime p,
>
>> There are an infinite number of odd primes p. What limits would you
>> suggest for the search: 3 <= p <= ???. Searching an infinite number
>> of primes can take an infinite amount of time.
>
> You start with p=3 and work up from there until the product of your
> primes that work is greater than the square root of the target.
>
>>> iterate through non-zero integers k until ...
|
| Show full article (3.33Kb) |
| no comments |
|
|
Author: NamehereNamehere Date: Jun 9, 2008 10:08
"Enrico" aol.com> wrote in message
news:b9973821-1509-4452-8d47-f2a1e6293b09@s33g2000pri.googlegroups.com...
On Jun 9, 8:11 am, JSH gmail.com> wrote:> On Jun 9, 3:57 am, rossum coldmail.com> wrote:
>>> On Sun, 8 Jun 2008 19:22:54 -0700 (PDT), JSH gmail.com> wrote:>>>After my latest miss-steps I went back, looked everything over again
>>>and finally noticed something that was bizarrely, simply obvious and
>>>staring me in the face.
>
>>>Here's the factoring algorithm that results.
>
>>>Given an odd target composite T, for a test odd prime p,
>
>> There are an infinite number of odd primes p. What limits would you
>> suggest for the search: 3 <= p <= ???. Searching an infinite number
>> of primes can take an infinite amount of time.
>
> You start with p=3 and work up from there until the product of your
> primes that work is greater than the square root of the target. ...
|
| Show full article (1.56Kb) |
| no comments |
|
|
|
|
|
Author: JSHJSH Date: Jun 9, 2008 16:52
On Jun 9, 7:09 am, Enrico aol.com> wrote:> On Jun 8, 8:22�pm, JSH gmail.com> wrote:
>
>
>>> After my latest miss-steps I went back, looked everything over again
>> and finally noticed something that was bizarrely, simply obvious and
>> staring me in the face.
>
>> Here's the factoring algorithm that results.
>
>> Given an odd target composite T, for a test odd prime p, iterate
>> through non-zero integers k until
>
>> T - k^2 mod p
>
>> is a quadratic residue. If no k will work, discard that prime p and
>> try another. �If one does work then get � from
>
>> �2 = (T - k^2)(k^2)^{-1} mod p. ...
|
| Show full article (3.30Kb) |
| no comments |
|
|
|
|
