Factorization theory wrong? Or algorithmic error?

On Apr 5, 6:19 pm, barker

Prove it. Demonstrate the factors.........

quasi

2012-04-06 00:07:29 UTC

On Fri, 06 Apr 2012 00:19:32 +0200, barker

Post by barker
11195609436169473474006154090025752843698874651430
10602130506309766179075300607267132230420289234876
95623178805395619821799868743856430058731438452818
437316840959014392166803390411010978334873
which tested algorithms suggest is prime, but which
I have factorized.

I seriously doubt that C is composite.

quasi

Pubkeybreaker

2012-04-05 23:25:21 UTC

Post by quasi
On Fri, 06 Apr 2012 00:19:32 +0200, barker

I seriously doubt that C is composite.
quasi

It is NOT composite. I ran it through my implementation of
APR-CL (Adleman, Pomerance, Rumely, Cohen, Lenstra)

barker

2012-04-07 01:48:10 UTC

I seriously doubt that C is composite.

It is NOT composite. I ran it through my implementation of
APR-CL (Adleman, Pomerance, Rumely, Cohen, Lenstra)

Thank you. I had already checked such results, though, naturally,
not with your implementation of the algorithms. I appreciate your
skepticism, as I experienced the same, and still, at some levels,
can't believe it. But multiplication is hard to argue against. I
have not found any other number besides "C" with the previously
mentioned property, by the way. I am trying to do so by generating
candidate "A"s (the 347 digit number that decomposed into just two
prime factors - or so the algorithms stated).

For the avoidance of doubt:

1) My original post <5365904fd6ca3b4f0811f9c0e9b00688> in this
thread contains no error or errors.

2) Within that post, I explained why I am delaying supplying the
factors, other than stating that the smaller one is almost 2^300.

3) You do probably know of me, though not because of access to
a computer capable of factorizing a ~1150 bit number or even a
~640 bit one ("C"). I am posting pseudonymously, and sufficiently
anonymously - check the headers, if you wish.

4) I suggest ignoring giganews poster CWatters <colin.watters@
turnersoak.plus.com>, of whom I have no knowledge but who appears
to be other than serious (by choosing mathwarehouse.com!).

I repeat - my post starting this thread contains no mistake(s).

Thank you,

"barker"

quasi

2012-04-07 03:31:56 UTC

On Sat, 07 Apr 2012 03:48:10 +0200, barker

I seriously doubt that C is composite.

It is NOT composite. I ran it through my implementation of
APR-CL (Adleman, Pomerance, Rumely, Cohen, Lenstra)

Thank you. I had already checked such results, though, naturally,
not with your implementation of the algorithms. I appreciate your
skepticism, as I experienced the same, and still, at some levels,
can't believe it. But multiplication is hard to argue against. I
have not found any other number besides "C" with the previously
mentioned property, by the way. I am trying to do so by generating
candidate "A"s (the 347 digit number that decomposed into just two
prime factors - or so the algorithms stated).
1) My original post <5365904fd6ca3b4f0811f9c0e9b00688> in this
thread contains no error or errors.
2) Within that post, I explained why I am delaying supplying the
factors, other than stating that the smaller one is almost 2^300.
3) You do probably know of me, though not because of access to
a computer capable of factorizing a ~1150 bit number or even a
~640 bit one ("C"). I am posting pseudonymously, and sufficiently
anonymously - check the headers, if you wish.
turnersoak.plus.com>, of whom I have no knowledge but who appears
to be other than serious (by choosing mathwarehouse.com!).
I repeat - my post starting this thread contains no mistake(s).
Thank you,
"barker"

It's hard to bet against someone who is presumably looking at
two easily veriable factors.

So at this point, I'll retract my "I seriously doubt it", but
I'm still at least somewhat skeptical.

quasi

quasi

2012-04-07 03:59:17 UTC

Post by quasi
On Sat, 07 Apr 2012 03:48:10 +0200, barker

I seriously doubt that C is composite.

It is NOT composite. I ran it through my implementation of
APR-CL (Adleman, Pomerance, Rumely, Cohen, Lenstra)

Thank you. I had already checked such results, though, naturally,
not with your implementation of the algorithms. I appreciate your
skepticism, as I experienced the same, and still, at some levels,
can't believe it. But multiplication is hard to argue against. I
have not found any other number besides "C" with the previously
mentioned property, by the way. I am trying to do so by generating
candidate "A"s (the 347 digit number that decomposed into just two
prime factors - or so the algorithms stated).
1) My original post <5365904fd6ca3b4f0811f9c0e9b00688> in this
thread contains no error or errors.
2) Within that post, I explained why I am delaying supplying the
factors, other than stating that the smaller one is almost 2^300.
3) You do probably know of me, though not because of access to
a computer capable of factorizing a ~1150 bit number or even a
~640 bit one ("C"). I am posting pseudonymously, and sufficiently
anonymously - check the headers, if you wish.
turnersoak.plus.com>, of whom I have no knowledge but who appears
to be other than serious (by choosing mathwarehouse.com!).
I repeat - my post starting this thread contains no mistake(s).
Thank you,
"barker"

It's hard to bet against someone who is presumably looking at
two easily verifiable factors.
So at this point, I'll retract my "I seriously doubt it", but
I'm still at least somewhat skeptical.

Assuming c really is composite, it's pretty clear that c is an
absolute Euler pseudoprime. If that's the case, I can understand
why some of the factoring algorithms might view c as prime.
What's strange is that if, as you say, c has a smaller prime factor
than a, why wouldn't the factoring algorithms find the smaller
prime factor first?

In any case, assuming you can generates such examples at will,
how do you propose to use your example-generating algorithm to
make money?

quasi

unruh

2012-04-07 14:35:27 UTC

["Followup-To:" header set to alt.security.pgp.]

Post by quasi
On Sat, 07 Apr 2012 03:48:10 +0200, barker

I seriously doubt that C is composite.

It is NOT composite. I ran it through my implementation of
APR-CL (Adleman, Pomerance, Rumely, Cohen, Lenstra)

Thank you. I had already checked such results, though, naturally,
not with your implementation of the algorithms. I appreciate your
skepticism, as I experienced the same, and still, at some levels,
can't believe it. But multiplication is hard to argue against. I
have not found any other number besides "C" with the previously
mentioned property, by the way. I am trying to do so by generating
candidate "A"s (the 347 digit number that decomposed into just two
prime factors - or so the algorithms stated).
1) My original post <5365904fd6ca3b4f0811f9c0e9b00688> in this
thread contains no error or errors.
2) Within that post, I explained why I am delaying supplying the
factors, other than stating that the smaller one is almost 2^300.
3) You do probably know of me, though not because of access to
a computer capable of factorizing a ~1150 bit number or even a
~640 bit one ("C"). I am posting pseudonymously, and sufficiently
anonymously - check the headers, if you wish.
turnersoak.plus.com>, of whom I have no knowledge but who appears
to be other than serious (by choosing mathwarehouse.com!).
I repeat - my post starting this thread contains no mistake(s).
Thank you,
"barker"

It's hard to bet against someone who is presumably looking at
two easily verifiable factors.
So at this point, I'll retract my "I seriously doubt it", but
I'm still at least somewhat skeptical.

Because it does not factor by starting at 1, trying to divide, on
failure incriment by 3 and try again. He would certainly not have
arrived at a 2^300 sized number yet 10^100=10^83 years at a ns per
division.

Post by quasi
In any case, assuming you can generates such examples at will,
how do you propose to use your example-generating algorithm to
make money?
quasi

William Hughes

2012-04-07 03:53:15 UTC

Post by quasi
On Sat, 07 Apr 2012 03:48:10 +0200, barker

I seriously doubt that C is composite.

It is NOT composite. I ran it through my implementation of
APR-CL (Adleman, Pomerance, Rumely, Cohen, Lenstra)

Thank you. I had already checked such results, though, naturally,
not with your implementation of the algorithms. I appreciate your
skepticism, as I experienced the same, and still, at some levels,
can't believe it. But multiplication is hard to argue against. I
have not found any other number besides "C" with the previously
mentioned property, by the way. I am trying to do so by generating
candidate "A"s (the 347 digit number that decomposed into just two
prime factors - or so the algorithms stated).
1) My original post <5365904fd6ca3b4f0811f9c0e9b00688> in this
thread contains no error or errors.
2) Within that post, I explained why I am delaying supplying the
factors, other than stating that the smaller one is almost 2^300.
3) You do probably know of me, though not because of access to
a computer capable of factorizing a ~1150 bit number or even a
~640 bit one ("C"). I am posting pseudonymously, and sufficiently
anonymously - check the headers, if you wish.
turnersoak.plus.com>, of whom I have no knowledge but who appears
to be other than serious (by choosing mathwarehouse.com!).
I repeat - my post starting this thread contains no mistake(s).
Thank you,
"barker"

It's hard to bet against someone who is presumably looking at
two easily veriable factors.

Not at all, it is easy to bet that an anonymous sci.math poster is
deluded or lying. Indeed register my bet that
"barker" will never provide two putative factors of C.
(However, he will not come up with any excuse even remotely as
entertaining as those of JSH)

- William Hughes

Post by quasi
So at this point, I'll retract my "I seriously doubt it", but
I'm still at least somewhat skeptical.
quasi

quasi

2012-04-07 18:45:19 UTC

Post by barker
I repeat - my post starting this thread contains no
mistake(s).

It's hard to bet against someone who is presumably looking
at two easily verifiable factors.

Not at all, it is easy to bet that an anonymous sci.math
poster is deluded or lying. Indeed register my bet that
"barker" will never provide two putative factors of C.
(However, he will not come up with any excuse even
remotely as entertaining as those of JSH)

Perhaps he actually _is_ JSH.

In any case, there have been enough independent confirmations
that C is prime that, at this point, it's clear that C really
_is_ prime, beyond any reasonable doubt.

Thus, as you say, the OP is either deluded or lying, and I
suspect the latter -- just another troll looking for attention.
Had I looked at the headers, I would have noticed that the OP
crossposted the original message to alt.politics (for which the
article is clearly off-topic), thus providing another indication
that the OP was trolling.

quasi

Tonico

2012-04-07 18:05:56 UTC

Post by barker
I repeat - my post starting this thread contains no
mistake(s).

It's hard to bet against someone who is presumably looking
at two easily verifiable factors.

Not at all, it is easy to bet that an anonymous sci.math
poster is deluded or lying. Indeed register my bet that
"barker" will never provide two putative factors of C.
(However, he will not come up with any excuse even
remotely as entertaining as those of JSH)

Perhaps he actually _is_ JSH.
In any case, there have been enough independent confirmations
that C is prime that, at this point, it's clear that C really
_is_ prime, beyond any reasonable doubt.
Thus, as you say, the OP is either deluded or lying, and I
suspect the latter -- just another troll looking for attention.
Had I looked at the headers, I would have noticed that the OP
crossposted the original message to alt.politics (for which the
article is clearly off-topic), thus providing another indication
that the OP was trolling.
quasi

Well, in the meantime this Barker character already succeeded to
extract 22 (and with this present one, 23) posts...:>)

I also don't believe he'll ever post any factors nor he'll confess he
made a mistake, but in the meanwhile he must be having some fun. Kudos
to him.

Tonio

unruh

2012-04-07 19:17:32 UTC

Post by barker
I repeat - my post starting this thread contains no
mistake(s).

It's hard to bet against someone who is presumably looking
at two easily verifiable factors.

Not at all, it is easy to bet that an anonymous sci.math
poster is deluded or lying. Indeed register my bet that
"barker" will never provide two putative factors of C.
(However, he will not come up with any excuse even
remotely as entertaining as those of JSH)

Perhaps he actually _is_ JSH.
In any case, there have been enough independent confirmations
that C is prime that, at this point, it's clear that C really
_is_ prime, beyond any reasonable doubt.
Thus, as you say, the OP is either deluded or lying, and I
suspect the latter -- just another troll looking for attention.
Had I looked at the headers, I would have noticed that the OP
crossposted the original message to alt.politics (for which the
article is clearly off-topic), thus providing another indication
that the OP was trolling.

Has the deterministic primality test been done on it, or only the
probabilistic ones?

Post by quasi
quasi

Edward A. Falk

2012-04-07 10:08:35 UTC

Post by quasi
It's hard to bet against someone who is presumably looking at
two easily veriable factors.

It's easy to bet against someone who claims to have made a
radical discovery in the field of mathematics but refuses to
back up their claim.

You'd hardly be giving up any huge secrets by proving that
you know the factors of that number. So why not post them?

--
-Ed Falk, ***@despams.r.us.com
http://thespamdiaries.blogspot.com/

r***@robert-earl-hazelett.com

2012-04-08 22:06:05 UTC

Post by quasi
On Fri, 06 Apr 2012 00:19:32 +0200, barker

C'mon . . . What's the matter with you guys? You're all scrambling
around like a bunch of monkeys trying to fuck a football. Don't you
remember that the proof is in the pudding? Or that this is a "put up
or shut up" world?

To determine if barker is a troll, all you have to do is create a
VERY LARGE number by multiplying two VERY LARGE primes. Invite him to
factor your number and, if he CANNOT or WILL NOT do so, he probably is
a harmless troll. Ignore him from that moment onward.

On the other hand, if he sends you the correct answer by return
mail you better sit up and take notice. Maybe even shed a few tears.
If he promptly sends the correct answer you can say goodbye to PGP and
GPG. One day after that your suspected troll will probably be a
millionaire. Surely NSA (or someone else) will reward him handsomely
for his solution or perhaps to keep his mouth shut about it. Of course
they might even decide to "snuff" him. One way or the other he'll be
out of your hair. :^)

Rob

Jan Andres

2012-04-08 23:15:14 UTC

Post by r***@robert-earl-hazelett.com

Post by quasi
On Fri, 06 Apr 2012 00:19:32 +0200, barker

C'mon . . . What's the matter with you guys? You're all scrambling
around like a bunch of monkeys trying to fuck a football.

Thank you so much for coming straight to the point.

Anyway, this one _was_ funny after all. I mean, hey, a troll whose
trolling consists of emulating a crank, coming up with something
apparently meant as an april fool's joke, except that it comes on Apr 5,
and it isn't funny.

How much more absurd can it get? :-)

Post by r***@robert-earl-hazelett.com
remember that the proof is in the pudding? Or that this is a "put up
or shut up" world?
To determine if barker is a troll, all you have to do is create a
VERY LARGE number by multiplying two VERY LARGE primes. Invite him to
factor your number and, if he CANNOT or WILL NOT do so, he probably is
a harmless troll. Ignore him from that moment onward.
On the other hand, if he sends you the correct answer by return
mail you better sit up and take notice. Maybe even shed a few tears.
If he promptly sends the correct answer you can say goodbye to PGP and
GPG. One day after that your suspected troll will probably be a
millionaire. Surely NSA (or someone else) will reward him handsomely
for his solution or perhaps to keep his mouth shut about it. Of course
they might even decide to "snuff" him. One way or the other he'll be
out of your hair. :^)
Rob

--
Jan Andres <***@gmx.net>

Edward A. Falk

2012-04-09 15:34:46 UTC

Post by r***@robert-earl-hazelett.com

Post by quasi
On Fri, 06 Apr 2012 00:19:32 +0200, barker

Ahhh, that's the problem with Usenet. There's no "like" button.

Well said.

--
-Ed Falk, ***@despams.r.us.com
http://thespamdiaries.blogspot.com/

CWatters

2012-04-06 16:49:49 UTC

-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA512
We (mathematicians) have grown to accept the primality checkers as
gospel. So did I, until recently.
This could be big, or it could be I've overlooked something, though I
have hunted for 3 days for a flaw. I'd appreciate if you could check
this over for me. This post is digitally signed in case I need to prove
ownership, should my discovery (if it is a discovery) be stolen.
As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 bits),
3634908448770161716619462884730373820150226880205007030541419827683585
7931761274740311086713549497603607279611408949613526779622187756741117
9048935484829402996681944342388178421558785023331981868685440034884277
9396792124395994336764804183754455993340622344242614470170379064513230
0552661368276733695867117608484513671228954258971153834928109857741
I won't tell you how I generated A, because if there's no flaw in what
I've done (I intend to make real money out of this, if it is possible),
the way I came up with A is a giveaway to the whole process.
I won't ask you to factorize A, because you may not be able to. Here is
3246726736489147307461784686107468324672673648914730746178468610746834
6821883878114173728372983219193183717113173468218838781141737283729832
1919318371711317
** that is, smaller as identified by all the factorization algorithms
that I have encountered. If you are not professional mathematicians
http://www.alpertron.com.ar/ECM.HTM
which will work on any modern web browser, to confirm what I have
just stated (i.e., that A is composite, B is prime and that A/B is an
integer; whether A/B is prime is moot).
ECM's author Dario Alpern has diligently implemented factorization
algorithms. His implementations are not in question (I assume they are
accurate, as do my colleagues) - it is the theory itself that is now
in question.
Divide A by B to get the 192 decimal digit number C. Since 192/2< 156,
it follows that if B was the smaller prime factor of A, then C must be
prime.
{Lemma: Assume C was non-prime. Then it must have at least one prime
factor that is less than 97 (= 192/2 + 1) decimal digits long. This
would falsify the algorithmic result that B, at 156 decimal digits, is
the smallest prime factor of A. Therefore C must be prime.}
I didn't want to give you C (= A/B) as I want you to (trivially) compute
it yourself (but for the lazy, it appears at the end of this post).
Now check C's primality. C should be prime, per the lemma above. Right?
Indeed, all the primality checkers I have tested show that C is prime.
http://www.alpertron.com.ar/ECM.HTM
Well, I can tell you that I have factorized C... and hand-checked it, as
at first I could not believe the fluke finding.
C's smaller factor is almost 2^300, so C's decomposition is non-trivial.
In the time window before you can brute-force this, I will disclose its
1) got me to A (Hint: diagonalization, Cantor), and
2) factorized C.
But at this point, I do not want to disclose C's factors, until I have
heard the more competent fellow mathematicians here confirm C's alleged
primality, according to the algorithms we all becomed conditioned to
believing are true.
I do hope I have not overlooked anything. Your assistance is appreciated.
Thank you,
"barker" (associate of the late falsified non-dullrich Dr Pertti Lounesto)
1119560943616947347400615409002575284369887465143010602130506309766179
0753006072671322304202892348769562317880539561982179986874385643005873
1438452818437316840959014392166803390411010978334873
which tested algorithms suggest is prime, but which I have factorized.
-----BEGIN PGP SIGNATURE-----
VQEcBAEBCgAGBQJOe7YRAAoJEAjjY4weks8oA7QIAK3ELb/+NKP1vLPT8f7HQTaf
Ym-qnG0TdO44RMJdbqpxsp6DoMx5JkMgluha8y6LIV3rBHHDKGQx3YwKzVTT5r81
DOOQ-r3LQdLgmoemhdot2Dse16XQ7OoWzvJw-qvvYYBZ0S/J2SsrAFUAoQAe35/4
9NkVg3-JSzV+AFPQyv5hpS780v0cObSPl7yz32MypgvZkYZupC3xP/3Pdl8Fg205
NkiDEaDl-JcIKM8ARJJtndd7cfNBKZ3Bh1OEQ1NwPFEMZ6uAR3S/DLdF0dY1MMxr
RRcluph+ML-mTRZngA8NG9qRCBQT2IgTZNatjnZv2pcwgC0MddUnyS07bNypHg8=
=87KU
-----END PGP SIGNATURE-----

For what it's worth a quick cut and paste produced..

http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-test.php?number=111956094361694734740061540900257528436988746514301060213050630976617907530060726713223042028923487695623178805395619821799868743856430058731438452818437316840959014392166803390411010978334873

Results of Prime Factorization test
111956094361694734740061540900257528436988746514301060213050630976617907530060726713223042028923487695623178805395619821799868743856430058731438452818437316840959014392166803390411010978334873
is not a prime number

Obviously if you ask it to look for factors it will time out.

Pubkeybreaker

2012-04-06 17:08:29 UTC

For what it's worth a quick cut and paste produced..
http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-te...
Results of Prime Factorization test
111956094361694734740061540900257528436988746514301060213050630976617907530060726713223042028923487695623178805395619821799868743856430058731438452818437316840959014392166803390411010978334873
is not a prime number
Obviously if you ask it to look for factors it will time out.- Hide quoted text -
- Show quoted text -

Another lying spammer. www.mathwarehouse.com does not have the
algorithms for even
testing primality of large integers. And the phrase "Prime
Factorization test" is nonsense.

I have not only tested this with my own software, but also with two
other independent pieces
of software. They all say that it is prime.

And the claim that the O.P. has somehow factored the number is crap
as well. I pretty much know
all of the people who have the ability to perform a factorization of a
192-digit composite.
And the O.P. isn't one of them.

William Hughes

2012-04-06 20:58:43 UTC

For what it's worth a quick cut and paste produced..
http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-te...
Results of Prime Factorization test
111956094361694734740061540900257528436988746514301060213050630976617907530 06072671322304202892348769562317880539561982179986874385643005873143845281 8437316840959014392166803390411010978334873
is not a prime number
Obviously if you ask it to look for factors it will time out.- Hide quoted text -
- Show quoted text -

Another lying spammer.

Never assume malice where incompetence is sufficient.
(It would appear that the site does in fact use a "Prime Factorization
test" for primality.
i.e. a number is prime iff it does not have any proper prime factors.
My guess is that
the applet calls the unix utility "factor")

- William Hughes

Pubkeybreaker

2012-04-06 23:47:17 UTC

For what it's worth a quick cut and paste produced..
http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-te...
Results of Prime Factorization test
111956094361694734740061540900257528436988746514301060213050630976617907530 06072671322304202892348769562317880539561982179986874385643005873143845281 8437316840959014392166803390411010978334873
is not a prime number
Obviously if you ask it to look for factors it will time out.- Hide quoted text -
- Show quoted text -

Another lying spammer.

If indeed it tests for primality by trial division and concludes that
a number is prime by
failure to find a factor even when it fails to take the division up to
sqrt(N)......
[i.e. I will assume that the algorithm looking for factors is indeed
incompetently done)

Since the number does NOT have any prime factors, the conclusion
therefore
should have been that the number is prime. Yet the spam poster claimed
that the
number was NOT prime. Therefore, if he tested for primality and
concluded
that the number was composite, he must have found a factor while
refusing to
tell what it is.

My conclusion of lying spammer still stands.

William Hughes

2012-04-07 00:27:10 UTC

For what it's worth a quick cut and paste produced..
http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-te...
Results of Prime Factorization test
111956094361694734740061540900257528436988746514301060213050630976617907530 06072671322304202892348769562317880539561982179986874385643005873143845281 8437316840959014392166803390411010978334873
is not a prime number
Obviously if you ask it to look for factors it will time out.- Hide quoted text -
- Show quoted text -

Another lying spammer.

If indeed it tests for primality by trial division and concludes that
a number is prime by
failure to find a factor even when it fails to take the division up to
sqrt(N)......
[i.e. I will assume that the algorithm looking for factors is indeed
incompetently done)
Since the number does NOT have any prime factors, the conclusion
therefore
should have been that the number is prime. Yet the spam poster claimed
that the
number was NOT prime. Therefore, if he tested for primality and
concluded
that the number was composite, he must have found a factor while
refusing to
tell what it is.
My conclusion of lying spammer still stands.

Nope. The applet is poorly designed. If you put in too large a
number it
returns "not prime" for anything. The applet will only handle 8
decimal
digits or so !!!!!! (I was obviously generous when I guessed it
called "factor"
in the background).

Jan Andres

2012-04-07 12:09:30 UTC

["Followup-To:" header set to sci.physics.]

I, too, have a homegrown primality prover I wrote a couple of years ago,
based on the classical methods ("n-1" and "n+1" tests as described near
the beginning of Crandall&Pomerance's Primality Proving section). I just
ran the number through it and, surprise, it also came out as a proven
prime.

quasi

2012-04-06 21:40:00 UTC

On Fri, 06 Apr 2012 17:49:49 +0100, CWatters

Post by CWatters
For what it's worth a quick cut and paste produced..
<http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization-calculator.php>
Results of Prime Factorization test
111956094361694734740061540900257528436988746514301060213050630976617907530060726713223042028923487695623178805395619821799868743856430058731438452818437316840959014392166803390411010978334873
is not a prime number

Obviously the applet cannot correctly handle numbers that large.

quasi

William Hughes

2012-04-06 20:46:50 UTC

On Apr 5, 7:19 pm, barker

-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA512
We (mathematicians) have grown to accept the primality checkers as
gospel. So did I, until recently.
This could be big, or it could be I've overlooked something, though I
have hunted for 3 days for a flaw. I'd appreciate if you could check
this over for me. This post is digitally signed in case I need to prove
ownership, should my discovery (if it is a discovery) be stolen.
As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 bits),
3634908448770161716619462884730373820150226880205007030541419827683585
7931761274740311086713549497603607279611408949613526779622187756741117
9048935484829402996681944342388178421558785023331981868685440034884277
9396792124395994336764804183754455993340622344242614470170379064513230
0552661368276733695867117608484513671228954258971153834928109857741
I won't tell you how I generated A, because if there's no flaw in what
I've done (I intend to make real money out of this, if it is possible),
the way I came up with A is a giveaway to the whole process.
I won't ask you to factorize A, because you may not be able to. Here is
3246726736489147307461784686107468324672673648914730746178468610746834
6821883878114173728372983219193183717113173468218838781141737283729832
1919318371711317
** that is, smaller as identified by all the factorization algorithms
that I have encountered. If you are not professional mathematicians
http://www.alpertron.com.ar/ECM.HTM
which will work on any modern web browser, to confirm what I have
just stated (i.e., that A is composite, B is prime and that A/B is an
integer; whether A/B is prime is moot).
ECM's author Dario Alpern has diligently implemented factorization
algorithms. His implementations are not in question (I assume they are
accurate, as do my colleagues) - it is the theory itself that is now
in question.
Divide A by B to get the 192 decimal digit number C. Since 192/2 < 156,
it follows that if B was the smaller prime factor of A, then C must be
prime.
{Lemma: Assume C was non-prime. Then it must have at least one prime
factor that is less than 97 (= 192/2 + 1) decimal digits long. This
would falsify the algorithmic result that B, at 156 decimal digits, is
the smallest prime factor of A. Therefore C must be prime.}
I didn't want to give you C (= A/B) as I want you to (trivially) compute
it yourself (but for the lazy, it appears at the end of this post).
Now check C's primality. C should be prime, per the lemma above. Right?
Indeed, all the primality checkers I have tested show that C is prime.
http://www.alpertron.com.ar/ECM.HTM
Well, I can tell you that I have factorized C... and hand-checked it, as
at first I could not believe the fluke finding.
C's smaller factor is almost 2^300, so C's decomposition is non-trivial.
In the time window before you can brute-force this, I will disclose its
1) got me to A (Hint: diagonalization, Cantor), and
2) factorized C.
But at this point, I do not want to disclose C's factors, until I have
heard the more competent fellow mathematicians here confirm C's alleged
primality, according to the algorithms we all becomed conditioned to
believing are true.
I do hope I have not overlooked anything. Your assistance is appreciated.
Thank you,
"barker" (associate of the late falsified non-dullrich Dr Pertti Lounesto)
1119560943616947347400615409002575284369887465143010602130506309766179
0753006072671322304202892348769562317880539561982179986874385643005873
1438452818437316840959014392166803390411010978334873
which tested algorithms suggest is prime, but which I have factorized.
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VQEcBAEBCgAGBQJOe7YRAAoJEAjjY4weks8oA7QIAK3ELb/+NKP1vLPT8f7HQTaf
Ym-qnG0TdO44RMJdbqpxsp6DoMx5JkMgluha8y6LIV3rBHHDKGQx3YwKzVTT5r81
DOOQ-r3LQdLgmoemhdot2Dse16XQ7OoWzvJw-qvvYYBZ0S/J2SsrAFUAoQAe35/4
9NkVg3-JSzV+AFPQyv5hpS780v0cObSPl7yz32MypgvZkYZupC3xP/3Pdl8Fg205
NkiDEaDl-JcIKM8ARJJtndd7cfNBKZ3Bh1OEQ1NwPFEMZ6uAR3S/DLdF0dY1MMxr
RRcluph+ML-mTRZngA8NG9qRCBQT2IgTZNatjnZv2pcwgC0MddUnyS07bNypHg8=
=87KU
-----END PGP SIGNATURE-----

I predict that if and when you provide your putative factors, one of
them will have a prime factor less that 1,000,000

- William Hughes

barker

2012-04-08 11:22:42 UTC

On Apr 5, 7:19=A0pm, barker

have hunted for 3 days for a flaw. I'd appreciate if you could chec=

this over for me. This post is digitally signed in case I need to p=

rove

ownership, should my discovery (if it is a discovery) be stolen.
As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 b=

its),

3634908448770161716619462884730373820150226880205007030541419827683=

585

7931761274740311086713549497603607279611408949613526779622187756741=

117

9048935484829402996681944342388178421558785023331981868685440034884=

277

9396792124395994336764804183754455993340622344242614470170379064513=

230

0552661368276733695867117608484513671228954258971153834928109857741
I won't tell you how I generated A, because if there's no flaw in w=

hat

I've done (I intend to make real money out of this, if it is possib=

le),

the way I came up with A is a giveaway to the whole process.
I won't ask you to factorize A, because you may not be able to. Her=

e is

its "smaller prime factor"** ("B"), which is 156 decimal digits lon=
3246726736489147307461784686107468324672673648914730746178468610746=

834

6821883878114173728372983219193183717113173468218838781141737283729=

832

1919318371711317
** that is, smaller as identified by all the factorization algorith=

that I have encountered. If you are not professional mathematicians
=A0http://www.alpertron.com.ar/ECM.HTM
which will work on any modern web browser, to confirm what I have
just stated (i.e., that A is composite, B is prime and that A/B is =

integer; whether A/B is prime is moot).
ECM's author Dario Alpern has diligently implemented factorization
algorithms. His implementations are not in question (I assume they =

are

accurate, as do my colleagues) - it is the theory itself that is no=

in question.
Divide A by B to get the 192 decimal digit number C. =A0Since 192/2=

< 156,

it follows that if B was the smaller prime factor of A, then C must=

prime.
{Lemma: Assume C was non-prime. Then it must have at least one prim=

factor that is less than 97 (=3D 192/2 + 1) decimal digits long. Th=

would falsify the algorithmic result that B, at 156 decimal digits,=

the smallest prime factor of A. Therefore C must be prime.}
I didn't want to give you C (=3D A/B) as I want you to (trivially) =

compute

it yourself (but for the lazy, it appears at the end of this post).
Now check C's primality. C should be prime, per the lemma above. Ri=

ght?

Indeed, all the primality checkers I have tested show that C is pri=

me.

=A0http://www.alpertron.com.ar/ECM.HTM
Well, I can tell you that I have factorized C... and hand-checked i=

t, as

at first I could not believe the fluke finding.
C's smaller factor is almost 2^300, so C's decomposition is non-tri=

vial.

In the time window before you can brute-force this, I will disclose=

its

1) got me to A (Hint: diagonalization, Cantor), and
2) factorized C.
But at this point, I do not want to disclose C's factors, until I h=

ave

heard the more competent fellow mathematicians here confirm C's all=

eged

primality, according to the algorithms we all becomed conditioned t=

believing are true.
I do hope I have not overlooked anything. Your assistance is apprec=

iated.

Thank you,
"barker" (associate of the late falsified non-dullrich Dr Pertti Lo=

unesto)

1119560943616947347400615409002575284369887465143010602130506309766=

179

0753006072671322304202892348769562317880539561982179986874385643005=

873

1438452818437316840959014392166803390411010978334873
which tested algorithms suggest is prime, but which I have factoriz=

ed.

I predict that if and when you provide your putative factors, one of
them will have a prime factor less that 1,000,000
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 - William Hughes

Since it is readily checked that C has no prime factors less than 1,000=

,000

(or even less than 1,000,000,000), your prediction amounts to barker
not revealing factors at all. =A0:)
hagman

=20
Nope. The above is only valid if the putative factors are in fact
factors of C.
Since C does not have any proper factors they will not be.
=20
- William Hughes

Well, at least I can write C as product of two integers such that neither
of these integers has a prime factor < 1,000,000.
Up to permutation of factors there's only one way to do so. ;)

At last, someone is at least a little bit on the right track?

Very poor reading and analytical skills shown in sci.math, no better than
in alt.politics.org.nsa. The original post contains no mistake(s),
error(s)
or wrong statement(s). Was it read carefully enough? Apparently, no.

I will supply a further hint. If we add together the 2 factors of C, to
arrive at D, the largest prime factor of D is the 183 decimal digit
number:
358050379911852746241673609972200391570944003061600841442151335677224869
764099592907918958324761500403496107209279988661853710490640317981357174
732801778565826180964783476648958068743

Thank you,

"barker" of Lowell, Mass.

Jan Andres

2012-04-08 13:08:38 UTC

On Apr 5, 7:19=A0pm, barker

have hunted for 3 days for a flaw. I'd appreciate if you could chec=

this over for me. This post is digitally signed in case I need to p=

rove

ownership, should my discovery (if it is a discovery) be stolen.
As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 b=

its),

3634908448770161716619462884730373820150226880205007030541419827683=

585

7931761274740311086713549497603607279611408949613526779622187756741=

117

9048935484829402996681944342388178421558785023331981868685440034884=

277

9396792124395994336764804183754455993340622344242614470170379064513=

230

0552661368276733695867117608484513671228954258971153834928109857741
I won't tell you how I generated A, because if there's no flaw in w=

hat

I've done (I intend to make real money out of this, if it is possib=

le),

the way I came up with A is a giveaway to the whole process.
I won't ask you to factorize A, because you may not be able to. Her=

e is

its "smaller prime factor"** ("B"), which is 156 decimal digits lon=
3246726736489147307461784686107468324672673648914730746178468610746=

834

6821883878114173728372983219193183717113173468218838781141737283729=

832

1919318371711317
** that is, smaller as identified by all the factorization algorith=

integer; whether A/B is prime is moot).
ECM's author Dario Alpern has diligently implemented factorization
algorithms. His implementations are not in question (I assume they =

are

accurate, as do my colleagues) - it is the theory itself that is no=

in question.
Divide A by B to get the 192 decimal digit number C. =A0Since 192/2=

< 156,

it follows that if B was the smaller prime factor of A, then C must=

prime.
{Lemma: Assume C was non-prime. Then it must have at least one prim=

factor that is less than 97 (=3D 192/2 + 1) decimal digits long. Th=

would falsify the algorithmic result that B, at 156 decimal digits,=

the smallest prime factor of A. Therefore C must be prime.}
I didn't want to give you C (=3D A/B) as I want you to (trivially) =

compute

it yourself (but for the lazy, it appears at the end of this post).
Now check C's primality. C should be prime, per the lemma above. Ri=

ght?

Indeed, all the primality checkers I have tested show that C is pri=

me.

=A0http://www.alpertron.com.ar/ECM.HTM
Well, I can tell you that I have factorized C... and hand-checked i=

t, as

at first I could not believe the fluke finding.
C's smaller factor is almost 2^300, so C's decomposition is non-tri=

vial.

In the time window before you can brute-force this, I will disclose=

its

1) got me to A (Hint: diagonalization, Cantor), and
2) factorized C.
But at this point, I do not want to disclose C's factors, until I h=

ave

heard the more competent fellow mathematicians here confirm C's all=

eged

primality, according to the algorithms we all becomed conditioned t=

believing are true.
I do hope I have not overlooked anything. Your assistance is apprec=

iated.

Thank you,
"barker" (associate of the late falsified non-dullrich Dr Pertti Lo=

unesto)

1119560943616947347400615409002575284369887465143010602130506309766=

179

0753006072671322304202892348769562317880539561982179986874385643005=

873

1438452818437316840959014392166803390411010978334873
which tested algorithms suggest is prime, but which I have factoriz=

ed.

I predict that if and when you provide your putative factors, one of
them will have a prime factor less that 1,000,000
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 - William Hughes

Since it is readily checked that C has no prime factors less than 1,000=

,000

(or even less than 1,000,000,000), your prediction amounts to barker
not revealing factors at all. =A0:)
hagman

=20
Nope. The above is only valid if the putative factors are in fact
factors of C.
Since C does not have any proper factors they will not be.
=20
- William Hughes

Well, at least I can write C as product of two integers such that neither
of these integers has a prime factor < 1,000,000.
Up to permutation of factors there's only one way to do so. ;)

At last, someone is at least a little bit on the right track?
Very poor reading and analytical skills shown in sci.math, no better than
in alt.politics.org.nsa. The original post contains no mistake(s),
error(s)
or wrong statement(s). Was it read carefully enough? Apparently, no.
I will supply a further hint. If we add together the 2 factors of C, to
arrive at D, the largest prime factor of D is the 183 decimal digit
358050379911852746241673609972200391570944003061600841442151335677224869
764099592907918958324761500403496107209279988661853710490640317981357174
732801778565826180964783476648958068743

It divides C+1. Bit late for april fool's day, huh? If you were trying
to be funny, you were't quite successful. And yes, your OP does contain
"wrong statement(s)". You claimed explicitly that your factorization was
"non-trivial" and that the smaller factor was almost 2^300.

Pertti's Ghost

2012-04-08 17:55:30 UTC

On 2012-04-08, barker

Post by barker
Very poor reading and analytical skills shown in sci.math, no better than
in alt.politics.org.nsa. The original post contains no mistake(s), error(s)
or wrong statement(s). Was it read carefully enough? Apparently, no.
I will supply a further hint. If we add together the 2 factors of C, to
arrive at D, the largest prime factor of D is the 183 decimal digit
358050379911852746241673609972200391570944003061600841442151335677224869
764099592907918958324761500403496107209279988661853710490640317981357174
732801778565826180964783476648958068743

It divides C+1.

For sure it does.

Bit late for april fool's day, huh?

I disagree it was late.

In barker's first post, which arrived at my news server on 5 Apr 2012
15:20 hrs,

This could be big, or it could be I've overlooked something, though I have
hunted for 3 days for a flaw.

Given that remailers were used, as he declared in his second post, and
their
chain latencies are in the order of a day or two, that could well imply he
started this voyage on April 1.

And yes, your OP does contain "wrong statement(s)".
You claimed explicitly that your factorization was "non-trivial"

"Non-trivial" is not defined, so its meaning lies in the eyes of its
beholder.
Further, any manipulation of a 640 bit number may well be considered to be
a
"non-trivial" operation.

and that the smaller factor was almost 2^300.

1 was the smaller factor. And 1 is almost 2^300, certainly by reference to
the
yardstick that A is larger than 2^1150. barker was careful not to say that
the
smaller factor was prime (at least 1 of the "contributors" to this thread
seems
to think 1 is prime).

I agree with barker, poor reading and analytical skills shown in sci.math,
where the term "factor" is read as if it implied "proper factor"... barker
never claimed C was composite (non-prime), just that he could factor it.
;D

barker 1, sci.math 0 ?

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

karl

2012-04-08 18:07:11 UTC

On 2012-04-08, barker

It divides C+1.

For sure it does.

Bit late for april fool's day, huh?

I disagree it was late.
In barker's first post, which arrived at my news server on 5 Apr 2012
15:20 hrs,

This could be big, or it could be I've overlooked something, though I have
hunted for 3 days for a flaw.

Given that remailers were used, as he declared in his second post, and
their
chain latencies are in the order of a day or two, that could well imply he
started this voyage on April 1.

And yes, your OP does contain "wrong statement(s)".
You claimed explicitly that your factorization was "non-trivial"

"Non-trivial" is not defined, so its meaning lies in the eyes of its
beholder.
Further, any manipulation of a 640 bit number may well be considered to be
a
"non-trivial" operation.

and that the smaller factor was almost 2^300.

1 was the smaller factor. And 1 is almost 2^300, certainly by reference to
the
yardstick that A is larger than 2^1150. barker was careful not to say that
the
smaller factor was prime (at least 1 of the "contributors" to this thread
seems
to think 1 is prime).
I agree with barker, poor reading and analytical skills shown in sci.math,
where the term "factor" is read as if it implied "proper factor"... barker
never claimed C was composite (non-prime), just that he could factor it.
;D
barker 1, sci.math 0 ?

This is from the first post of barker:

As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 bits),
which I'll call A:

Seems as if you are wrong?

Pertti's Ghost

2012-04-09 23:17:30 UTC

As it is not Sunday, I will break with my tradition of not engaging with

Post by barker
As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 bits),
Seems as if you are wrong?

You have confused yourself between A and C.

Of course A is composite, and barker never claimed otherwise. Quoting
from barker's post:

This is A, a positive integer:

3634908448770161716619462884730373820150226880205007030541419827683585
7931761274740311086713549497603607279611408949613526779622187756741117
9048935484829402996681944342388178421558785023331981868685440034884277
9396792124395994336764804183754455993340622344242614470170379064513230
0552661368276733695867117608484513671228954258971153834928109857741

This is B, a positive integer:

3246726736489147307461784686107468324672673648914730746178468610746834
6821883878114173728372983219193183717113173468218838781141737283729832
1919318371711317

This is C, a positive integer:

1119560943616947347400615409002575284369887465143010602130506309766179
0753006072671322304202892348769562317880539561982179986874385643005873
1438452818437316840959014392166803390411010978334873

And it is trivial to check:

A = B x C

Therefore, A is composite. barker was correct.

C, not A, is the number about which barker wrote:

"which tested algorithms suggest is prime, but which I have factorized."

but barker never wrote these factors were prime factors (1 is not
prime) or proper factors (C is not a proper factor of C), and his
claim that C's smaller factor was "almost 2^300" is, of course, not
false, as 1 is almost 2^300 in the scale of other numbers mentioned,
e.g., A > 2^1150 and C > 2^635 >> (2^300)^2 >> 2^300.

barker's factorization was C = 1 x C.

Everything barker wrote was correct. His critics missed every single
finesse. Even after their errors are corrected, they keep arguing.

Sadly-
any idiot can post to sci.math, and
many do.

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

Richard Tobin

2012-04-10 00:20:21 UTC

Post by Pertti's Ghost
but barker never wrote these factors were prime factors (1 is not
prime) or proper factors (C is not a proper factor of C),

If they are not proper factors, then he is a troll, and that's as
bad as being a liar.

Post by Pertti's Ghost
and his
claim that C's smaller factor was "almost 2^300" is, of course, not
false, as 1 is almost 2^300 in the scale of other numbers mentioned,

And so, it seems, are you.

-- Richard

unruh

2012-04-10 00:26:08 UTC

Post by Pertti's Ghost
As it is not Sunday, I will break with my tradition of not engaging with

Post by barker
As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 bits),
Seems as if you are wrong?

You have confused yourself between A and C.

Sounds like we know who Barker is.

Post by Pertti's Ghost
Of course A is composite, and barker never claimed otherwise. Quoting
3634908448770161716619462884730373820150226880205007030541419827683585
7931761274740311086713549497603607279611408949613526779622187756741117
9048935484829402996681944342388178421558785023331981868685440034884277
9396792124395994336764804183754455993340622344242614470170379064513230
0552661368276733695867117608484513671228954258971153834928109857741
3246726736489147307461784686107468324672673648914730746178468610746834
6821883878114173728372983219193183717113173468218838781141737283729832
1919318371711317
1119560943616947347400615409002575284369887465143010602130506309766179
0753006072671322304202892348769562317880539561982179986874385643005873
1438452818437316840959014392166803390411010978334873
A = B x C
Therefore, A is composite. barker was correct.
"which tested algorithms suggest is prime, but which I have factorized."
but barker never wrote these factors were prime factors (1 is not
prime) or proper factors (C is not a proper factor of C), and his
claim that C's smaller factor was "almost 2^300" is, of course, not
false, as 1 is almost 2^300 in the scale of other numbers mentioned,
e.g., A > 2^1150 and C > 2^635 >> (2^300)^2 >> 2^300.

The scale for almost is given in the sentence itself, 2^300. Not 2^635
oe 2^1150.

Post by Pertti's Ghost
barker's factorization was C = 1 x C.

You know this how? You are barker?

Post by Pertti's Ghost
Everything barker wrote was correct. His critics missed every single
finesse. Even after their errors are corrected, they keep arguing.

No, it was not. I am afraid he strayed beyond the bounds where pilpul
turns into lies.

Post by Pertti's Ghost
Sadly-
any idiot can post to sci.math, and
many do.

Yes, they do, don't they. (and sci.physics, and
alt.security.pgp,and comp.security.pgp.discuss, and
alt.politics.org.nsa)

Tonico

2012-04-10 02:06:03 UTC

Post by Pertti's Ghost
As it is not Sunday, I will break with my tradition of not engaging with

....................................................

Post by Pertti's Ghost
Sadly-
any idiot can post to sci.math, and
many do.

Oh, my boy! What an accurate sentence and how nice of you to serve as
an example to it.

Thanx

Tonico

2012-04-08 18:13:35 UTC

On 2012-04-08, barker

It divides C+1.

For sure it does.

Bit late for april fool's day, huh?

I disagree it was late.
In barker's first post, which arrived at my news server on 5 Apr 2012

This could be big, or it could be I've overlooked something, though I have
hunted for 3 days for a flaw.

Given that remailers were used, as he declared in his second post, and
their
chain latencies are in the order of a day or two, that could well imply he
started this voyage on April 1.

And yes, your OP does contain "wrong statement(s)".
You claimed explicitly that your factorization was "non-trivial"

"Non-trivial" is not defined, so its meaning lies in the eyes of its
beholder.
Further, any manipulation of a 640 bit number may well be considered to be
a
"non-trivial" operation.

and that the smaller factor was almost 2^300.

1 was the smaller factor. And 1 is almost 2^300, certainly by reference to
the
yardstick that A is larger than 2^1150. barker was careful not to say that
the
smaller factor was prime (at least 1 of the "contributors" to this thread
seems
to think 1 is prime).
I agree with barker, poor reading and analytical skills shown in sci.math,
where the term "factor" is read as if it implied "proper factor"... barker
never claimed C was composite (non-prime), just that he could factor it.
;D
barker 1, sci.math 0 ?

First, you seem to think (figure of speech) that there was a contest
somewhere. There wasn't.

Second, "to factor a number" has a definite meaning in mathematics,
and writing it as itself times 1 does not fit it (can you say why?),
but even if it did you seem way to joyful for what you believe is a
cunning joke, which perhaps could have been. It wasn't though, and
either you're barker him/herself or his/her jester assistant.

Tonio

unruh

2012-04-08 21:30:21 UTC

On 2012-04-08, barker

It divides C+1.

For sure it does.

Bit late for april fool's day, huh?

I disagree it was late.
In barker's first post, which arrived at my news server on 5 Apr 2012
15:20 hrs,

This could be big, or it could be I've overlooked something, though I have
hunted for 3 days for a flaw.

Given that remailers were used, as he declared in his second post, and
their
chain latencies are in the order of a day or two, that could well imply he
started this voyage on April 1.

He may have started HIS voyage on Apr 1 but he wrote and sent out the
post much later than Apr 1. He WAS late. And his post contained lies as
has been pointed out.
Even the sentence you quote "I may have overlooked something". If he
meant that 1 is a divisor, then that is hardly something he overlooked.
He knew it and posted with malice aforethought. So that is lie. And his
first is also a lie, since he knew it was not big. And he presented only
those two lies as the possible options.
And other lies have been pointed out.

And yes, your OP does contain "wrong statement(s)".
You claimed explicitly that your factorization was "non-trivial"

"Non-trivial" is not defined, so its meaning lies in the eyes of its
beholder.

No it does not. That "1" is a divisor is trivial under all definitions
of trivial.

Post by Pertti's Ghost
Further, any manipulation of a 640 bit number may well be considered to be
a
"non-trivial" operation.

But dividing by 1 is not.

and that the smaller factor was almost 2^300.

Again, 1 is not almost 2^300. It differs from 2^300 by a factor of
2^300.

Post by Pertti's Ghost
I agree with barker, poor reading and analytical skills shown in sci.math,
where the term "factor" is read as if it implied "proper factor"... barker
never claimed C was composite (non-prime), just that he could factor it.
;D

And now you are making up terms. Where do you get the term "proper
factor" from.

I am sorry, it is not due to the poor reading skills of the readers, but
poor writing skills of the writer.

Post by Pertti's Ghost
barker 1, sci.math 0 ?

Pertti's Ghost

2012-04-09 01:45:27 UTC

As it is Sunday, I will break with my tradition of not engaging with or

On 2012-04-08, barker

It divides C+1.

For sure it does.

Bit late for april fool's day, huh?

I disagree it was late.
In barker's first post, which arrived at my news server on 5 Apr 2012
15:20 hrs,

This could be big, or it could be I've overlooked something, though I have
hunted for 3 days for a flaw.

Given that remailers were used, as he declared in his second post, and
their
chain latencies are in the order of a day or two, that could well imply he
started this voyage on April 1.

He may have started HIS voyage on Apr 1 but he wrote and sent out the
post much later than Apr 1. He WAS late.

The messages were sent, as the headers show, via slow->bunker->cripto, out
via frell judging by the MID. Total remailer latency a couple of days.

Frelled Mixmaster Stats - 2012-04-01
Pinger circuit #2 Stats-Version: 2.0
Generated: Sun 01 Apr 2012 08:20:01 GMT
Mixmaster Latent-Hist Latent Uptime-Hist Uptime Options
------------------------------------------------------------------------
banana 000000000000 :05 ++++++++++++ 100.0% DPR IN1
austria 001010110112 :26 ++++++++++++ 100.0% R GO ATLE IN9
dizum 221010121010 :29 ++++++++++++ 100.0% PR GO ATLEUIN0
winters 212212132222 1:21 ++++++++++++ 100.0% D R IN
pobox 324229422221 1:30 ++++++++++++ 100.0% D R IN
freierede 1222362A2232 1:38 ++++++++++++ 100.0% DPR GO ATLE IN
cmeclax 239122322213 1:40 ++++++++++++ 100.0% D R GO ATLE IN
awxcnx 225422322232 1:40 ++++++++++++ 100.0% DPR GO ATLE IN9
anon 221727B17212 1:42 ++++++++++++ 100.0% DPR GO ATLE IN9
hermetix 222122234632 1:42 ++++++++++++ 100.0% DPR GO ATLE IN9
eurovibes ????????7572 1:49 ????????++++ 100.0% PR IN
foo 244135264424 1:56 ++++++++++++ 100.0% DPR IN
lulunga 242262323234 2:27 ++++++++++++ 100.0% DPR IN
cripto DCA7AABBA121 10:09 +++++++++++9 99.8% PR GO ATLE IN1
frell 854D87454A81 4:33 +++++++++++9 99.4% PR GO ATLEUIN9
rabbi 22B272222321 1:28 +++++++++++7 98.4% D R IN
kroken 121111222231 :49 +++++++++++7 98.1% D R GO TLEUIN9
kulin 000000021014 :08 +++++++76++5 88.9% DPRHG XATLEUIN9
slow BBCG57HC2CC? 27:14 ++++++++6756 83.7% D R IN1
nickserv 43214235564? 2:06 6+7332+64+50 53.7% D R UIN
bunker ????4?3????9 3:28 000020200005 10.0% DPR UIN

The third column gives latencies

Post by unruh
And his post contained lies as has been pointed out.

Lies? Not even one. Not even a false statement. But then, you are not a
mathematician, as you have been brought into sci.math only by the
xposting.
You probably think the statement that:
"every even prime numbers greater than 2 is an odd number"
is other than true.

Post by unruh
Even the sentence you quote "I may have overlooked something". If he
meant that 1 is a divisor, then that is hardly something he overlooked.
He knew it and posted with malice aforethought.

There, we agree.

Post by unruh
So that is lie.

Wrong again.

Post by unruh
And his first is also a lie, since he knew it was not big.

A non-mathematician, aren't you? To you, 1 is not big.
Try comparing it to zero. :)

And yes, your OP does contain "wrong statement(s)".
You claimed explicitly that your factorization was "non-trivial"

"Non-trivial" is not defined, so its meaning lies in the eyes of its
beholder.

No it does not. That "1" is a divisor is trivial under all definitions
of trivial.

Post by Pertti's Ghost
Further, any manipulation of a 640 bit number may well be considered to be
a "non-trivial" operation.

But dividing by 1 is not.

Any manipulation of a 640 bit number may well be considered to be a
"non-trivial" operation. Read until understood, unruh. I understand you
are trolling, and I am being indulgent even by responding. :)

and that the smaller factor was almost 2^300.

Again, 1 is not almost 2^300. It differs from 2^300 by a factor of
2^300.

First, "almost" is undefined and thus true by default. Heard of Zermelo-
Fraenkel?

Second, let's run with your """logic""", in the interest of indulgence.

It is true that 2^300 is 2^300 times bigger than 2^0.

2^1150 (in fact, a bit more than that, A) was the size comparator, quoted
at the beginning.

And it is also true that 2^1150 is 2^850 times bigger than 2^300.

Therefore, 1 is 2^550 times closer to 2^300 than 2^300 is to 2^1150.

Now, 2^550 is a very, very big number. So 1 is immensely closer to 2^300
than either of them is to A.

Get it? "Close" is relative, and is not defined anyway.

And now you are making up terms. Where do you get the term "proper
factor" from.

Where do you get the term "factor" from? Each is well known:
https://en.wikipedia.org/wiki/Proper_factors

A "proper factor" or "proper divisor" of integer C is a positive integral
factor of C which is less than C. If C has no proper factors other than 1,
then and only then is C prime. By definition, 1 is not prime (and not
non-prime either).

This is the problem permitting trash^H^H^H^H^Hgenii from sci.physics to
post here in sci.math...

The O.P., barker of Mass., never wrote or suggested that C was non-prime.
He merely wrote or implied that he had factorized it. Which he had.
C = 1 x C, and 1, in the context of the number C (let alone A!) is fairly
described as almost 2^300.

Had he written that he had decomposed C into proper factors, or prime
factors, he would have been wrong. C is a prime factor but not a proper
factor, and 1 is a proper factor but not a prime factor.

And no doubt lying too, as he would have known he was wrong.

But, he did not write that. He simply wrote that he had factorized C.

Post by unruh
I am sorry, it is not due to the poor reading skills of the readers, but
poor writing skills of the writer.

You are quite stupid. I conclude that as hitherto suspected you, like
JSH of FLT infamy, are from sci.physics. No surprise, then.

Post by Pertti's Ghost
barker 1, sci.math 0 ?

And barker 2, sci.physics 0, I think.

The truly stupid don't realize, even after it has been explained to them,
that without a misstatement being made, they were had.

Post by unruh
To determine if barker is a troll, all you have to do is create a
VERY LARGE number by multiplying two VERY LARGE primes. Invite him to
factor your number

I am 100% confident that barker would have replied that he was only able
to factorize A (the 1150+ bit number that was "nearly prime", i.e., had
exactly two, distinct prime factors) because of the special nature of A
that made C appear to be prime to all the test-engines while in fact he
had factorized it, and therefore could not replicate this with other
nearly prime numbers of A's magnitude.

Occam's Razor suggests that while barker presented A, B and C to you in
that sequence, he arrived at them with A at the end of the sequence.

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

William Hughes

2012-04-09 04:02:38 UTC

Post by Pertti's Ghost
Any manipulation of a 640 bit number may well be considered to be a
"non-trivial" operation.

The exact quote is

"C's decomposition is non-trivial"

Certainly, the decomposition C=C*1 is trivial (note it does not
require any manipulation of C).

-William Hughes

William Hughes

2012-04-09 14:50:28 UTC

Post by Pertti's Ghost
Any manipulation of a 640 bit number may well be considered to be a
"non-trivial" operation.

The exact quote is
"C's decomposition is non-trivial"

Of course the full quote is

"C's smaller factor is almost 2^300,
so C's decomposition is non-trivial.

This has the form A implies B, and in formal logic
"A implies B" is true when both A and B are false.
Thus the OP's claim, all the statements are true.

However, the normal reading of the full quote is

"C's smaller factor is almost 2^300 and this implies
C's decomposition is non-trivial"

This is false.

-William Hughes

karl

2012-04-09 17:59:45 UTC

You did not read my answer and the quotation there?
Maybe you are just ig..

Pertti's Ghost

2012-04-08 19:36:03 UTC

On 2012-04-08, barker

It divides C+1.

For sure it does.

Bit late for april fool's day, huh?

I disagree it was late.

In barker's first post, which arrived at my news server on 5 Apr 2012
15:20 hrs,

This could be big, or it could be I've overlooked something, though I have
hunted for 3 days for a flaw.

Given that remailers were used, as he declared in his second post, and
their
chain latencies are in the order of a day or two, that could well imply he
started this voyage on April 1.

And yes, your OP does contain "wrong statement(s)".
You claimed explicitly that your factorization was "non-trivial"

"Non-trivial" is not defined, so its meaning lies in the eyes of its
beholder.
Further, any manipulation of a 640 bit number may well be considered to be
a
"non-trivial" operation.

and that the smaller factor was almost 2^300.

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

Pubkeybreaker

2012-04-08 13:13:50 UTC

On Apr 8, 7:22 am, barker

On Apr 5, 7:19=A0pm, barker

have hunted for 3 days for a flaw. I'd appreciate if you could chec=

this over for me. This post is digitally signed in case I need to p=

rove

ownership, should my discovery (if it is a discovery) be stolen.
As part of my research into improving factorization algorithms, I
encountered this composite number, 347 decimal digits long (>1150 b=

its),

3634908448770161716619462884730373820150226880205007030541419827683=

585

7931761274740311086713549497603607279611408949613526779622187756741=

117

9048935484829402996681944342388178421558785023331981868685440034884=

277

9396792124395994336764804183754455993340622344242614470170379064513=

230

0552661368276733695867117608484513671228954258971153834928109857741
I won't tell you how I generated A, because if there's no flaw in w=

hat

I've done (I intend to make real money out of this, if it is possib=

le),

the way I came up with A is a giveaway to the whole process.
I won't ask you to factorize A, because you may not be able to. Her=

e is

its "smaller prime factor"** ("B"), which is 156 decimal digits lon=
3246726736489147307461784686107468324672673648914730746178468610746=

834

6821883878114173728372983219193183717113173468218838781141737283729=

832

1919318371711317
** that is, smaller as identified by all the factorization algorith=

integer; whether A/B is prime is moot).
ECM's author Dario Alpern has diligently implemented factorization
algorithms. His implementations are not in question (I assume they =

are

accurate, as do my colleagues) - it is the theory itself that is no=

in question.
Divide A by B to get the 192 decimal digit number C. =A0Since 192/2=

< 156,

it follows that if B was the smaller prime factor of A, then C must=

prime.
{Lemma: Assume C was non-prime. Then it must have at least one prim=

factor that is less than 97 (=3D 192/2 + 1) decimal digits long. Th=

would falsify the algorithmic result that B, at 156 decimal digits,=

the smallest prime factor of A. Therefore C must be prime.}
I didn't want to give you C (=3D A/B) as I want you to (trivially) =

compute

it yourself (but for the lazy, it appears at the end of this post).
Now check C's primality. C should be prime, per the lemma above. Ri=

ght?

Indeed, all the primality checkers I have tested show that C is pri=

me.

=A0http://www.alpertron.com.ar/ECM.HTM
Well, I can tell you that I have factorized C... and hand-checked i=

t, as

at first I could not believe the fluke finding.
C's smaller factor is almost 2^300, so C's decomposition is non-tri=

vial.

In the time window before you can brute-force this, I will disclose=

its

1) got me to A (Hint: diagonalization, Cantor), and
2) factorized C.
But at this point, I do not want to disclose C's factors, until I h=

ave

heard the more competent fellow mathematicians here confirm C's all=

eged

primality, according to the algorithms we all becomed conditioned t=

believing are true.
I do hope I have not overlooked anything. Your assistance is apprec=

iated.

Thank you,
"barker" (associate of the late falsified non-dullrich Dr Pertti Lo=

unesto)

1119560943616947347400615409002575284369887465143010602130506309766=

179

0753006072671322304202892348769562317880539561982179986874385643005=

873

1438452818437316840959014392166803390411010978334873
which tested algorithms suggest is prime, but which I have factoriz=

ed.

I predict that if and when you provide your putative factors, one of
them will have a prime factor less that 1,000,000
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 - William Hughes

Since it is readily checked that C has no prime factors less than 1,000=

,000

(or even less than 1,000,000,000), your prediction amounts to barker
not revealing factors at all. =A0:)
hagman

=20
Nope. The above is only valid if the putative factors are in fact
factors of C.
Since C does not have any proper factors they will not be.
=20
- William Hughes

Well, at least I can write C as product of two integers such that neither
of these integers has a prime factor < 1,000,000.
Up to permutation of factors there's only one way to do so. ;)

You wrote:

"Well, I can tell you that I have factorized C... and hand-checked it,
as
at first I could not believe the fluke finding.

C's smaller factor is almost 2^300, so C's decomposition is non-
trivial.
In the time window before you can brute-force this, I will disclose
its
factors, and the methods that:
1) got me to A (Hint: diagonalization, Cantor), and
2) factorized C. "

Since C is prime, any intelligent reader is going to conclude that you
are a spamming and lying piece of shit.

Post by barker
I will supply a further hint. If we add together the 2 factors of C, to
arrive at D, the largest prime factor of D is the 183 decimal digit
358050379911852746241673609972200391570944003061600841442151335677224869
764099592907918958324761500403496107209279988661853710490640317981357174
732801778565826180964783476648958068743

IMPOSSIBLE by the statement you made before. You stated that the
smallest
factor of C was approximately 2^300. The sum of the two factors of C
is
therefore approximately 2^300 + C/2^300 and this number is

Pubkeybreaker

2012-04-08 13:17:02 UTC

Post by Pubkeybreaker
On Apr 8, 7:22 am, barker