## RSA 576

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• Corresponding to my new factoring method, I believe that I have the first ten digits (not the last) of each factor of RSA576. Any one who wants to participate
Message 1 of 29 , Jan 13, 2002
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Corresponding to my new factoring method, I believe
that I have the first ten digits (not the last) of each
factor of RSA576.

Any one who wants to participate or contribute
program skills, should feel free to contact me at

miltbrown@...

Milton L. Brown

[Non-text portions of this message have been removed]
• Milton Brown wrote ... RSA576 was constructed with no regard to base 10. So if you can find an appreciable number of first digits of factors in that
Message 2 of 29 , Jan 13, 2002
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Milton Brown wrote

> I believe that I have the first ten digits (not the last)
> of each factor of RSA576.

RSA576 was constructed with no regard to base 10.
So if you can find an appreciable number of first
digits of factors in that miscellaneous base, Milton,
you ought to be able to find them in any base.

since specifying the initial digits of
a factor in enough bases would be enough
to nail it down.

Or at least it seems that way to me.

• Is there a paper or method on how to nail it down? I have read, but I forget where: ... if you know the first digits of an RSA factor then you can find the
Message 3 of 29 , Jan 13, 2002
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Is there a paper or method on how to "nail it" down?

I have read, but I forget where:

"... if you know the first digits of an RSA factor
then you can find the remaining digits."

Milton
----- Original Message -----
Sent: Sunday, January 13, 2002 4:37 PM

> Milton Brown wrote
>
> > I believe that I have the first ten digits (not the last)
> > of each factor of RSA576.
>
> RSA576 was constructed with no regard to base 10.
> So if you can find an appreciable number of first
> digits of factors in that miscellaneous base, Milton,
> you ought to be able to find them in any base.
>
> since specifying the initial digits of
> a factor in enough bases would be enough
> to nail it down.
>
> Or at least it seems that way to me.
>
>
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
• ... I doubt it. Assume that knowing the first ten digits of the prime factors allowed you to know the factors. Then one could try all possible combinations
Message 4 of 29 , Jan 13, 2002
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Milton Brown wrote:
>
> Is there a paper or method on how to "nail it" down?
>
> I have read, but I forget where:
>
> "... if you know the first digits of an RSA factor
> then you can find the remaining digits."
>
> Milton

I doubt it. Assume that knowing the first ten digits of the
prime factors allowed you to know the factors. Then one could
try all possible combinations of first ten digits until one
found the factors. Factoring is harder than that.

There is no known efficient factoring algorithm which computes
factors a digit at a time (or by groups of digits). That being
said, it hasn't been proven impossible either.

If one could find the units digits of the factors, in many bases,
one could find the complete factors using the Chinese Remainder
Theorem.

Have you verified your new factoring method on known numbers?
It seems a bit strange to go after RSA576; how can you check

Jack
• Jack Brennen wrote ... That was my feeling, too. David
Message 5 of 29 , Jan 13, 2002
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Jack Brennen wrote
> If one could find the units digits of the factors,
> in many bases, one could find the complete factors
> using the Chinese Remainder Theorem.
That was my feeling, too.
David
• Apologies to Milton and Jack: I think I was rather stupid. Knowing the leading N percent of the digits in one base is the same as knowing that percentage in
Message 6 of 29 , Jan 13, 2002
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Apologies to Milton and Jack:
I think I was rather stupid.
Knowing the leading N percent of the digits in one base
is the same as knowing that percentage in any other.
So unlike the trailing digits, little info
Sorry...
David
• Thanks for the note. I have used my method on RSA 100, 120, and 130. And, can upload these results if any one is interested. Milton L. Brown PS. I do remember
Message 7 of 29 , Jan 13, 2002
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Thanks for the note.

I have used my method on RSA 100, 120, and 130.

And, can upload these results if any one is interested.

Milton L. Brown

PS. I do remember reading the quote, I think it was Math of Comp
but could not find it in a quick search this weekend.

Is it possible to use ECM specifing the first digits of the factor?
----- Original Message -----
From: "Jack Brennen" <jack@...>
Sent: Sunday, January 13, 2002 5:57 PM
Subject: Re: [PrimeNumbers] Re: RSA 576

> Milton Brown wrote:
> >
> > Is there a paper or method on how to "nail it" down?
> >
> > I have read, but I forget where:
> >
> > "... if you know the first digits of an RSA factor
> > then you can find the remaining digits."
> >
> > Milton
>
> I doubt it. Assume that knowing the first ten digits of the
> prime factors allowed you to know the factors. Then one could
> try all possible combinations of first ten digits until one
> found the factors. Factoring is harder than that.
>
> There is no known efficient factoring algorithm which computes
> factors a digit at a time (or by groups of digits). That being
> said, it hasn't been proven impossible either.
>
> If one could find the units digits of the factors, in many bases,
> one could find the complete factors using the Chinese Remainder
> Theorem.
>
>
> Have you verified your new factoring method on known numbers?
> It seems a bit strange to go after RSA576; how can you check
>
> Jack
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
• ... If anyone is interested? I imagine that if your method is valid, that the whole mathematical community will be _very_ interested! Please do upload your
Message 8 of 29 , Jan 13, 2002
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Milton Brown wrote:
>
> I have used my method on RSA 100, 120, and 130.
>
> And, can upload these results if any one is interested.
>

If anyone is interested? I imagine that if your method is
valid, that the whole mathematical community will be _very_
interested!

Jack
• ... Could you explain this statement in more detail? Thanks. Robert Sitton Software Engineer & Digital Holographer rsitton@austin.rr.com ICQ: 28332295 ...
Message 9 of 29 , Jan 13, 2002
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> If one could find the units digits of the factors,
> in many bases, one could find the complete factors
> using the Chinese Remainder Theorem.

Could you explain this statement in more detail? Thanks.

Robert Sitton
Software Engineer & Digital Holographer
rsitton@...
ICQ: 28332295

-----Original Message-----
Sent: Sunday, January 13, 2002 8:17 PM

Jack Brennen wrote
> If one could find the units digits of the factors,
> in many bases, one could find the complete factors
> using the Chinese Remainder Theorem.
That was my feeling, too.
David

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• ... The units digits are represent the value reduced modulo the base of representation. So if I said I had a number which ended with a 3 in base 10, and
Message 10 of 29 , Jan 14, 2002
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On Sun, 13 January 2002, "Robert Sitton" wrote:
> > If one could find the units digits of the factors,
> > in many bases, one could find the complete factors
> > using the Chinese Remainder Theorem.
>
> Could you explain this statement in more detail? Thanks.

The units digits are represent the value reduced modulo the base of representation.
So if I said I had a number which ended with a '3' in base 10, and ended in a '18' base 21, you'd know that my number was 123 mod 210. Keep adding new bases, coprime to previous ones until you know the value mod something larger than the size of the number, and you've uniquely defined it.

Where did 123 come from above?

21 is 1 mod 10, and the inverse of 1 is 1 mod 10
So we have 1*21 is 1 mod 10, and 0 mod 21
Then 63 = 3*1*21 is 3 mod 10, and 0 mod 21

Similarly
10 is 10 mod 21 and the inverse of 10 is -2, or 19 mod 21
So we have 19*10 is 0 mod 10, and 1 mod 21
Then 3420=18*19*10 is 0 mod 10, and 18 mod 21

So 3483=63+3420 is 3 mod 10, and 18 mod 21
But the answer is only defined mod LCM(21,10)=210
3483 mod 210 == 123 mod 210.

Things are a bit messier with non-coprime bases, (i.e. you can end up with situations that contradict, and have no solution).
However, that's the guts of the CRT. A quick google for 'Chinese Remainder Theorem' should yield a fair few results.

HTH,
Phil

Don't be fooled, CRC Press are _not_ the good guys.
They've taken Wolfram's money - _don't_ give them yours.
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• ... Given purely on what Milton has already said, your gut feel is alright David, as long as you can make some assumtions about the method. For example, it
Message 11 of 29 , Jan 14, 2002
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On Sun, 13 January 2002, "djbroadhurst" wrote:
> Milton Brown wrote
>
> > I believe that I have the first ten digits (not the last)
> > of each factor of RSA576.
>
> RSA576 was constructed with no regard to base 10.
> So if you can find an appreciable number of first
> digits of factors in that miscellaneous base, Milton,
> you ought to be able to find them in any base.
>
> since specifying the initial digits of
> a factor in enough bases would be enough
> to nail it down.
>
> Or at least it seems that way to me.

Given purely on what Milton has already said, your gut feel is alright David, as long as you can make some assumtions about the method.
For example, it relies on being able to perform Milton's technique to arbitrary bases, and the effective computability of find small (<10 digit, say) multiples of large (>N-10 digit) prime (or other) powers which intersect the current range. For every such number you can find that intersects the current range, you'll gain on average 1 bit of information as you chop the range.
Of course it also assumes that the range information that is derivable in each base is different.

Two big assumtions, but we're not curently given much to base deductions on, are we. Having said that, perhaps we should just stop hypothesising until we can actually see what's on the slab.

Phil

Don't be fooled, CRC Press are _not_ the good guys.
They've taken Wolfram's money - _don't_ give them yours.
http://mathworld.wolfram.com/erics_commentary.html

Find the best deals on the web at AltaVista Shopping!
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• Since we re on the subject of RSA 576 ... Is there any news on the project to factor RSA 576 using GNFS? There was some polynomial selection activity last
Message 12 of 29 , Jan 14, 2002
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Since we're on the subject of RSA 576 ...

Is there any news on the project to factor RSA 576 using GNFS?
There was some polynomial selection activity last year, to which I
contributed, but it all seems to have gone very quiet recently.

I'm currently working on trying to implement Montgomery's square root
algorithm (for my own interest and education as much as anything),
and I've got most of the building blocks in place (thanks to Cohen's
extremely useful book). One of the steps in the square root algorithm
is to generate an integral basis for the number field given by the
polynomial, and to do this (using the Round2 algorithm) you need to
factor the discriminant of the polynomial. I tried throwing a
polynomial I found for RSA576 at Round2 to see what happened, and it
turns out that the discriminant contains a 181-digit composite which
doesn't fall to ECM easily; this is more digits than RSA576 itself!
Anyway, my question is: can we expect that we can find a "good"
polynomial for RSA576 whose discriminant can be factored more easily
than RSA576. If not, then is GNFS in trouble?

Chris

(sorry Phil, if you got this directly - I meant to post to the group)
• Milton, I would love to contribute, and will if you can tell me the first digits of the prime factors of the following smaller number:
Message 13 of 29 , Jan 14, 2002
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Milton,

I would love to contribute, and will if you can tell me the first digits of
the prime factors of the following smaller number:

13992260829028575833221309068175950734\
36466314335287375914323790498211182117\
84073926343628502967503672629

Regards,

Paul.

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• As I see it my method could be combined with the GNFS, which searches the congruence x^2 = y^2 mod N But, we can limit the search because we know the beginning
Message 14 of 29 , Jan 14, 2002
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As I see it my method could be combined with
the GNFS, which searches the congruence

x^2 = y^2 mod N

But, we can limit the search because we know the
beginning digits of x and y.

Attached are my computations for RSA100.

Milton L. Brown
miltbrown@...

----- Original Message -----
From: "ctcard_hotmail_com" <ctcard@...>
Sent: Monday, January 14, 2002 1:58 AM

> Since we're on the subject of RSA 576 ...
>
> Is there any news on the project to factor RSA 576 using GNFS?
> There was some polynomial selection activity last year, to which I
> contributed, but it all seems to have gone very quiet recently.
>
> I'm currently working on trying to implement Montgomery's square root
> algorithm (for my own interest and education as much as anything),
> and I've got most of the building blocks in place (thanks to Cohen's
> extremely useful book). One of the steps in the square root algorithm
> is to generate an integral basis for the number field given by the
> polynomial, and to do this (using the Round2 algorithm) you need to
> factor the discriminant of the polynomial. I tried throwing a
> polynomial I found for RSA576 at Round2 to see what happened, and it
> turns out that the discriminant contains a 181-digit composite which
> doesn't fall to ECM easily; this is more digits than RSA576 itself!
> Anyway, my question is: can we expect that we can find a "good"
> polynomial for RSA576 whose discriminant can be factored more easily
> than RSA576. If not, then is GNFS in trouble?
>
> Chris
>
> (sorry Phil, if you got this directly - I meant to post to the group)
>
>
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>

[Non-text portions of this message have been removed]
• Milton Brown wrote ... but yahoo removed the attachment. David
Message 15 of 29 , Jan 14, 2002
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Milton Brown wrote
> Attached are my computations for RSA100.
but yahoo removed the attachment.
David
• ... Fear not, said Phil (for mighty dread had siezed their troubled mind)! Milton sent me a copy directly, and I ve uploaded it to the files area.
Message 16 of 29 , Jan 15, 2002
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On Mon, 14 January 2002, "Milton Brown" wrote:
> As I see it my method could be combined with
> the GNFS, which searches the congruence
>
> x^2 = y^2 mod N
>
> But, we can limit the search because we know the
> beginning digits of x and y.
>
> Attached are my computations for RSA100.

Fear not, said Phil (for mighty dread had siezed their troubled mind)! Milton sent me a copy directly, and I've uploaded it to the files area. http://groups.yahoo.com/group/primenumbers/files/FactoringPrograms/rsa100.xls

I stuck it in the 'Programs' area as I hope it contains some actual algorithms and calculations rather than just a list of hand-entered numbers. However, it's in one of Billy-boy's proprietory and ever-changing file formats so I did nothing more than the following myself:

<<<
bash-2.05a\$ strings /tmp/rsa100.xls | more
Brown
B
"\$"#,##0_);\("\$"#,##0\)
"\$"#,##0_);[Red]\("\$"#,##0\)
"\$"#,##0.00_);\("\$"#,##0.00\)
"\$"#,##0.00_);[Red]\("\$"#,##0.00\)
_("\$"* #,##0_);_("\$"* \(#,##0\);_("\$"* "-"_);_(@_)
_(* #,##0_);_(* \(#,##0\);_(* "-"_);_(@_)
_("\$"* #,##0.00_);_("\$"* \(#,##0.00\);_("\$"* "-"??_);_(@_)
_(* #,##0.00_);_(* \(#,##0.00\);_(* "-"??_);_(@_)
Sheet1
Sheet2
Sheet3
implies
other
is 379..
RSA-100 factored by Manasse and Lenstra in 19912
379752279369436739228088727554456278545655366381993
40094690950920881030683735292761468389214899724061
0122963258952897654000350692006139E
N=152260502792253336053561837813263742971806811496138068865790849458\
and so on
MbP?_
DINU"
MbP?_
MbP?_
Brown
Brown
Microsoft Excel
Brown
Sheet1
Sheet2
Sheet3
Worksheets
>>>

So I'm as yet none the wiser, alas. However, the rest of you can have a browse now. I look forward to your enlightenment.

Phil

Don't be fooled, CRC Press are _not_ the good guys.
They've taken Wolfram's money - _don't_ give them yours.
http://mathworld.wolfram.com/erics_commentary.html

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• Phil, .xls extensions are used for Excel spreadsheets. Phew, these Unix kiddies, they know nowt about proper computing ;-). Regards, Paul. ...
Message 17 of 29 , Jan 15, 2002
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Phil,

.xls extensions are used for Excel spreadsheets. Phew, these Unix kiddies,
they know nowt about proper computing ;-).

Regards,

Paul.

> -----Original Message-----
> From: Phil Carmody [mailto:fatphil@...]
> Sent: 15 January 2002 08:58
> Subject: Re: [PrimeNumbers] Re: RSA 576
>
>
> On Mon, 14 January 2002, "Milton Brown" wrote:
> > As I see it my method could be combined with
> > the GNFS, which searches the congruence
> >
> > x^2 = y^2 mod N
> >
> > But, we can limit the search because we know the
> > beginning digits of x and y.
> >
> > Attached are my computations for RSA100.
>
> troubled mind)! Milton sent me a copy directly, and I've
> uploaded it to the files area.
> rams/rsa100.xls
>
> I stuck it in the 'Programs' area as I hope it contains some
> actual algorithms and calculations rather than just a list of
> hand-entered numbers. However, it's in one of Billy-boy's
> proprietory and ever-changing file formats so I did nothing
> more than the following myself:
>
> <<<
> bash-2.05a\$ strings /tmp/rsa100.xls | more
> Brown
>
> B
> "\$"#,##0_);\("\$"#,##0\)
> "\$"#,##0_);[Red]\("\$"#,##0\)
> "\$"#,##0.00_);\("\$"#,##0.00\)
> "\$"#,##0.00_);[Red]\("\$"#,##0.00\)
> _("\$"* #,##0_);_("\$"* \(#,##0\);_("\$"* "-"_);_(@_)
> _(* #,##0_);_(* \(#,##0\);_(* "-"_);_(@_)
> _("\$"* #,##0.00_);_("\$"* \(#,##0.00\);_("\$"* "-"??_);_(@_)
> _(* #,##0.00_);_(* \(#,##0.00\);_(* "-"??_);_(@_)
> Sheet1
> Sheet2
> Sheet3
> implies
> other
> is 379..
> RSA-100 factored by Manasse and Lenstra in 19912
> 379752279369436739228088727554456278545655366381993
> 40094690950920881030683735292761468389214899724061
> 0122963258952897654000350692006139E
> N=152260502792253336053561837813263742971806811496138068865790849458\
> and so on
> MbP?_
> DINU"
> MbP?_
> MbP?_
> Brown
> Brown
> Microsoft Excel
> Brown
> Sheet1
> Sheet2
> Sheet3
> Worksheets
> >>>
>
> So I'm as yet none the wiser, alas. However, the rest of you
> can have a browse now. I look forward to your enlightenment.
>
> Phil
>
> Don't be fooled, CRC Press are _not_ the good guys.
> They've taken Wolfram's money - _don't_ give them yours.
> http://mathworld.wolfram.com/erics_commentary.html
>
>
> Find the best deals on the web at AltaVista Shopping!
> http://www.shopping.altavista.com
>
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> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/

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• ... Maybe so, but much better informed 8-) ... Perhaps the material below might help. I apologise that it isn t perfect, but it might be a little more
Message 18 of 29 , Jan 15, 2002
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> troubled mind)! Milton sent me a copy directly, and I've

> I stuck it in the 'Programs' area as I hope it contains some
> actual algorithms and calculations rather than just a list of
> hand-entered numbers. However, it's in one of Billy-boy's
> proprietory and ever-changing file formats so I did nothing
> more than the following myself:

> So I'm as yet none the wiser, alas. However, the rest of you

Maybe so, but much better informed 8-)

> can have a browse now. I look forward to your enlightenment.

Perhaps the material below might help. I apologise that it isn't perfect, but it might be a little more readable. For the cognoscenti. I loaded the file into Excel and did a straightforward cut&paste into a "Plain Text" format email in Outlook.

No, I don't understand the computations here either. Some additional explanation would be valuable please, Milton. To be really helpful, could you perform and explain the corresponding manipulations for RSA-576?

Paul

=========================================================================

RSA-100 factored by Manasse and Lenstra in 1991

N=152260502792253336053561837813263742971806811496138068865790849458\
0122963258952897654000350692006139

2 4 30 17 4008 100 Breadth Best Search
3 17 31 30 4009 9
4 2 32 85 4010 6
5 49 33 10 4011 83
34 69 ↓ 3795 53 37975 49 379750 49
35 29 implies 3796 9 37976 2 379751 10
36 63 other 3797 5 379752 21
37 133 <= is 379.. 3798 35 379755 40 379753 54
38 5 379754 31
39 84 <= 377 15 379755 49
40 2 378 76 One factor is 3797… *
41 88 <= 379 59 and the other factor is 4009… 37975227 21
42 34 380 5 37975228 3
381 86 37975229 44

37975227936943673922808872755445627854565536638199 * 37975227…
and so on
40094690950920881030683735292761468389214899724061
• ... I think the distinction between intermediate working and algorithm needs to be drawn. Milton - there s _no_ algorithm in evidence yet. I believe that
Message 19 of 29 , Jan 15, 2002
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On Tue, 15 January 2002, "Paul Leyland" wrote:
> Perhaps the material below might help. I apologise that it isn't perfect, but it might be a little more readable. For the cognoscenti. I loaded the file into Excel and did a straightforward cut&paste into a "Plain Text" format email in Outlook.
>
>
> No, I don't understand the computations here either. Some additional explanation would be valuable please, Milton. To be really helpful, could you perform and explain the corresponding manipulations for RSA-576?

I think the distinction between "intermediate working" and "algorithm" needs to be drawn. Milton - there's _no_ algorithm in evidence yet. I believe that neither Euclid nor al-Kwarizmi could program any computer language at all, so it's nothing to do with coding as such. Just tell us what you're doing. Only then we can provide feedback.

> 3 17 31 30 4009 9
> 4 2 32 85 4010 6
> 5 49 33 10 4011 83
> 34 69 ?� 3795 53 37975 49 379750 49
> 35 29 implies 3796 9 37976 2 379751 10
> 36 63 other 3797 5 379752 21
> 37 133 <= is 379.. 3798 35 379755 40 379753 54
> 38 5 379754 31
> 39 84 <= 377 15 379755 49

Looks like a nightmare (with respawning, obviously) game of 3-dimensional minesweeper!

Phil

It is against US Department of Agriculture regulations to
advertise or sell as "Prime Rib" any cut of meat containing a
non-prime number of ribs.

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• ... indicates that I don t known it s an Excel spreadsheet? I thought not. The incentive to download Sun s Star Office still hasn t reached the positive side
Message 20 of 29 , Jan 15, 2002
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On Tue, 15 January 2002, "Paul Jobling" wrote:
> Phil,
>
> .xls extensions are used for Excel spreadsheets. Phew, these Unix kiddies,
> they know nowt about proper computing ;-).

And can you tell me which precise bit of:
> > However, it's in one of Billy-boy's
> > proprietory and ever-changing file formats
indicates that I don't known it's an Excel spreadsheet?

I thought not.

The incentive to download Sun's Star Office still hasn't reached the positive side of the number line.

Phil

It is against US Department of Agriculture regulations to
advertise or sell as "Prime Rib" any cut of meat containing a
non-prime number of ribs.

Find the best deals on the web at AltaVista Shopping!
http://www.shopping.altavista.com
• Yes, thanks. The numbers are from the residual function R= x * y - N , x and y bracketing primes which I have described in these notes before. The search
Message 21 of 29 , Jan 15, 2002
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Yes, thanks.

The numbers are from the residual function

R= x' * y' - N , x' and y' bracketing primes

which I have described in these notes before. The
search is for consistent minimums between the
two factors.

There is no complicated algorithm which you must divine!

I hope to explain this in a paper soon.

Milton L. Brown
miltbrown@...

----- Original Message -----
From: "Phil Carmody" <fatphil@...>
Sent: Tuesday, January 15, 2002 2:39 AM
Subject: RE: [PrimeNumbers] Re: RSA 576

> On Tue, 15 January 2002, "Paul Leyland" wrote:
> > Perhaps the material below might help. I apologise that it isn't
perfect, but it might be a little more readable. For the cognoscenti. I
loaded the file into Excel and did a straightforward cut&paste into a "Plain
Text" format email in Outlook.
> >
> >
> > No, I don't understand the computations here either. Some additional
you perform and explain the corresponding manipulations for RSA-576?
>
>
> I think the distinction between "intermediate working" and "algorithm"
needs to be drawn. Milton - there's _no_ algorithm in evidence yet. I
believe that neither Euclid nor al-Kwarizmi could program any computer
language at all, so it's nothing to do with coding as such. Just tell us
what you're doing. Only then we can provide feedback.
>
> > 3 17 31 30 4009 9
> > 4 2 32 85 4010 6
> > 5 49 33 10 4011 83
> > 34 69 ?« 3795 53
37975 49 379750 49
> > 35 29 implies 3796 9
37976 2 379751 10
> > 36 63 other 3797 5
379752 21
> > 37 133 <= is 379.. 3798
35 379755 40 379753 54
> > 38 5
379754 31
> > 39 84 <= 377 15
379755 49
>
>
> Looks like a nightmare (with respawning, obviously) game of 3-dimensional
minesweeper!
>
> Phil
>
> It is against US Department of Agriculture regulations to
> advertise or sell as "Prime Rib" any cut of meat containing a
> non-prime number of ribs.
>
>
>
> Find the best deals on the web at AltaVista Shopping!
> http://www.shopping.altavista.com
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
• PS: Concerning RSA576, I could explain them and give you the numbers for free, but why should I? A pot of gold waits at the end of that rainbow. Milton L.
Message 22 of 29 , Jan 15, 2002
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PS:

Concerning RSA576, I could explain them and give you

A pot of gold waits at the end of that rainbow.

Milton L. Brown
----- Original Message -----
From: "Phil Carmody" <fatphil@...>
Sent: Tuesday, January 15, 2002 2:39 AM
Subject: RE: [PrimeNumbers] Re: RSA 576

> On Tue, 15 January 2002, "Paul Leyland" wrote:
> > Perhaps the material below might help. I apologise that it isn't
perfect, but it might be a little more readable. For the cognoscenti. I
loaded the file into Excel and did a straightforward cut&paste into a "Plain
Text" format email in Outlook.
> >
> >
> > No, I don't understand the computations here either. Some additional
you perform and explain the corresponding manipulations for RSA-576?
>
>
> I think the distinction between "intermediate working" and "algorithm"
needs to be drawn. Milton - there's _no_ algorithm in evidence yet. I
believe that neither Euclid nor al-Kwarizmi could program any computer
language at all, so it's nothing to do with coding as such. Just tell us
what you're doing. Only then we can provide feedback.
>
> > 3 17 31 30 4009 9
> > 4 2 32 85 4010 6
> > 5 49 33 10 4011 83
> > 34 69 ?« 3795 53
37975 49 379750 49
> > 35 29 implies 3796 9
37976 2 379751 10
> > 36 63 other 3797 5
379752 21
> > 37 133 <= is 379.. 3798
35 379755 40 379753 54
> > 38 5
379754 31
> > 39 84 <= 377 15
379755 49
>
>
> Looks like a nightmare (with respawning, obviously) game of 3-dimensional
minesweeper!
>
> Phil
>
> It is against US Department of Agriculture regulations to
> advertise or sell as "Prime Rib" any cut of meat containing a
> non-prime number of ribs.
>
>
>
> Find the best deals on the web at AltaVista Shopping!
> http://www.shopping.altavista.com
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
• ... Fair enough. I ll be racing you to the factors, it appears. Anyone want a side bet on whether Milton or The Cabal get there first? Because I must declare
Message 23 of 29 , Jan 15, 2002
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> PS:
>
> Concerning RSA576, I could explain them and give you
>
> A pot of gold waits at the end of that rainbow.

Fair enough. I'll be racing you to the factors, it appears.

Anyone want a side bet on whether Milton or The Cabal get there first?
Because I must declare an interest, I must also decline to take part in
any side bets.

If The Cabal wins, my portion of the pot of gold will be donated to
charity, as have previous pots.

Here is a proposal to enable you to stake your claim to priority and yet
give nothing away. Write a small note containing the first few digits of
each factor, the first six would be easily enough to convince me. Take
that file and encrypt it with whatever crypto package you feel can't be
broken by the likes of me in less time than it would take me to factor
RSA-576. Post the result, or leave it on your web page, or somehow make
it publically available in good time. Make sure that you keep a copy of
the crypto key so that the public encrypted copy of your note can be
decrypted. If The Cabal, or anyone else other than you, publishes the
complete factorization before you do, you then reveal the crypto key so
that anyone can check your prediction. If the prediction is correct, I
can guarantee that a lot of people will be very interested in your
algorithm.

In effect: prove your claim, but do it safely.

Paul

Paul
• Paul Leyland asked ... I ll put 3 pints of organic on the Cabal. It seemed to me that if Milton has a method for determining leading digits of factors, it
Message 24 of 29 , Jan 19, 2002
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> Anyone want a side bet on whether Milton or The Cabal
> get there first?

I'll put 3 pints of organic on the Cabal.

It seemed to me that if Milton has a method for
determining leading digits of factors, it should
be applicable in any base.

As a matter of fact, I was able to determine the first
10 digits of each of the factors of RSA-130, in base 7,
as follows:

2231161355...
2461042421...

Perhaps my method is comparable to Milton's?

Be that as it may, I claim that mine could
have been applied at any time subsequent to
12 Apr 1996 13:49:43 EDT

• So, you have aligned with The Cabal. Even more formidable ! Not at 13:49:42 EDT ? ... From: djbroadhurst To:
Message 25 of 29 , Jan 19, 2002
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So, you have aligned with The Cabal.

Even more formidable !

Not at 13:49:42 EDT ?
----- Original Message -----
Sent: Saturday, January 19, 2002 5:05 PM

>
> > Anyone want a side bet on whether Milton or The Cabal
> > get there first?
>
> I'll put 3 pints of organic on the Cabal.
>
> It seemed to me that if Milton has a method for
> determining leading digits of factors, it should
> be applicable in any base.
>
> As a matter of fact, I was able to determine the first
> 10 digits of each of the factors of RSA-130, in base 7,
> as follows:
>
> 2231161355...
> 2461042421...
>
> Perhaps my method is comparable to Milton's?
>
> Be that as it may, I claim that mine could
> have been applied at any time subsequent to
> 12 Apr 1996 13:49:43 EDT
>
>
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
• now that the factors are known, were you right? if yes, i would like to know how your algorithm works. thommy
Message 26 of 29 , Dec 7, 2003
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now that the factors are known, were you right?
if yes, i would like to know how your algorithm works.

thommy

--- In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
wrote:
> Corresponding to my new factoring method, I believe
> that I have the first ten digits (not the last) of each
> factor of RSA576.
>
> Any one who wants to participate or contribute
> program skills, should feel free to contact me at
>
> miltbrown@e...
>
> Milton L. Brown
>
>
> [Non-text portions of this message have been removed]
• Please see message 5183.As you can see,a smart person could think that it might not be wise to make fool of himself very soon..so he gave the first digits of
Message 27 of 29 , Dec 7, 2003
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Please see message 5183.As you can see,a smart person
could think that it might not be wise to make fool of
himself very soon..so he "gave" the first digits of
RSA704 only.Milton's "algorithm" has only a finite
duration time..Wait till the factorization of RSA-704!

--- yummie_55555 <moonwar@...> wrote:
> now that the factors are known, were you right?
> if yes, i would like to know how your algorithm
> works.
>
> thommy
>
> --- In primenumbers@yahoogroups.com, "Milton Brown"
> <miltbrown@e...>
> wrote:
> > Corresponding to my new factoring method, I
> believe
> > that I have the first ten digits (not the last) of
> each
> > factor of RSA576.
> >
> > Any one who wants to participate or contribute
> > program skills, should feel free to contact me at
> >
> > miltbrown@e...
> >
> > Milton L. Brown
> >
> >
> > [Non-text portions of this message have been
> removed]
>
>

__________________________________
Do you Yahoo!?
http://photos.yahoo.com/
• Are you working in this area? I would be glad to work with you. Milton L. Brown miltbrown@earthlink.net ... From: yummie_55555 [mailto:moonwar@gmx.de] Sent:
Message 28 of 29 , Dec 7, 2003
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Are you working in this area?

I would be glad to work with you.

Milton L. Brown
miltbrown@...

-----Original Message-----
From: yummie_55555 [mailto:moonwar@...]
Sent: Sunday, December 07, 2003 2:30 AM

now that the factors are known, were you right?
if yes, i would like to know how your algorithm works.

thommy

--- In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
wrote:
> Corresponding to my new factoring method, I believe
> that I have the first ten digits (not the last) of each
> factor of RSA576.
>
> Any one who wants to participate or contribute
> program skills, should feel free to contact me at
>
> miltbrown@e...
>
> Milton L. Brown
>
>
> [Non-text portions of this message have been removed]

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
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• Hello all, I couldn t find my post on RSA-576, but I did, just now apply my method to RSA 576 With the following results for the larger factor First digits: 5
Message 29 of 29 , Dec 7, 2003
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Hello all,

I couldn't find my post on RSA-576, but
I did, just now apply my method to RSA 576
With the following results for the larger factor
First digits:

5
43 => 4
473 => 47
4725 => 472
47273 => 4727

This uses the Nyquist criteria to pin down the current digit.
I can provide the worksheets to anyone interest.

Hope this helps.

Milton L. Brown
miltbrown@...

-----Original Message-----
From: Pavlos S [mailto:pavlos199@...]
Sent: Sunday, December 07, 2003 2:45 AM
Subject: Re: [PrimeNumbers] Re: RSA 576

Please see message 5183.As you can see,a smart person
could think that it might not be wise to make fool of
himself very soon..so he "gave" the first digits of
RSA704 only.Milton's "algorithm" has only a finite
duration time..Wait till the factorization of RSA-704!

--- yummie_55555 <moonwar@...> wrote:
> now that the factors are known, were you right?
> if yes, i would like to know how your algorithm
> works.
>
> thommy
>
> --- In primenumbers@yahoogroups.com, "Milton Brown"
> <miltbrown@e...>
> wrote:
> > Corresponding to my new factoring method, I
> believe
> > that I have the first ten digits (not the last) of
> each
> > factor of RSA576.
> >
> > Any one who wants to participate or contribute
> > program skills, should feel free to contact me at
> >
> > miltbrown@e...
> >
> > Milton L. Brown
> >
> >
> > [Non-text portions of this message have been
> removed]
>
>

__________________________________
Do you Yahoo!?