mersenneforum.org

mersenneforum.org (https://www.mersenneforum.org/index.php)
-   Homework Help (https://www.mersenneforum.org/forumdisplay.php?f=78)
-   -   Many "Zeros" in Public Key, factorization easy ? (https://www.mersenneforum.org/showthread.php?t=12848)

Unregistered 2009-12-11 08:03

Many "Zeros" in Public Key, factorization easy ?
 
Hey all,

our Professor gave us a riddle concerning RSA, prime factorization. It roughly sounds like this:

When creating my public key, I was careless, the public key looks like this:

[CODE]
n = 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000004943449893861185811563806587897209039394765349228600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000004438724091153735385330145100740518448292725876750765167357855909678843140444732487940078664422013
[/CODE]

I am quite sure that the factorization is fairly easy on this one. But why ? Just cannot figure this one out :-( Would be great if someone could give me a tip on this one, does it have to do something with the factorial primes ?

Thank you very much,

Greetings

cheesehead 2009-12-11 11:49

What are n's particular traits? What particular characteristics of RSA public keys do you know, in general?

Unregistered 2009-12-11 12:34

n = (10^m+a)*(10^m+b) = pq
a*b = 4438724091153735385330145100740518448292725876750765167357855909678843140444732487940078664422013
a+b = 49434498938611858115638065878972090393947653492286
(should have the actual answer for you shortly...)
J

fivemack 2009-12-11 12:55

? A={big number}
? T=round(sqrt(A))
? issquare(T^2-A)
1
? A%round(T-sqrt(T^2-A))
0

Unregistered 2009-12-11 13:30

I'm actually having problems with this...
msieve gives the factorization of (a*b) as:
p1 factor: 3
p1 factor: 3
p2 factor: 31
p6 factor: 140551
p7 factor: 1691101
prp41 factor: 53192383744805292572364285928936275866737
prp43 factor: 1258348708567569132243547867193996784363281
Now, since a*b is 97 digits and a+b is 50 digits, then a and b must each be 49 digits (?).
Also, since a*b ends in 3, and a+b ends in 6, then a ends in 7, and b in 9 (?).
So far, so good - but I can't seem to arrange the factors above to give desired a,b....
Can anyone else help?
J

cheesehead 2009-12-11 13:31

[quote=Unregistered;198487]n = (10^m+a)*(10^m+b) = pq
a*b = 4438724091153735385330145100740518448292725876750765167357855909678843140444732487940078664422013
a+b = 49434498938611858115638065878972090393947653492286
[/quote]Check your assumptions.

Unregistered 2009-12-11 13:37

[QUOTE=cheesehead;198495]Check your assumptions.[/QUOTE]

You tell me! :)
J

Unregistered 2009-12-11 15:08

prp43=p43
J

[CODE]
Last login: Fri Dec 11 14:45:57 on ttys000
Desmond:~ james$ cd math
Desmond:math james$ cd gmp-ecpp
Desmond:gmp-ecpp james$ ./atkin49.gmp*
total = 3183
max = 111763

PI = 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914127372458701
******************
E = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901157383418793070215408914993488416750924476146066808226480016847741185374234544243710753907774499206955170227
******************
NATLOGONEPOINTNINE = 0.64185388617239472924480336140742337708875986273577482323572661701981349350622067937977027835051711557447795163484234193531054347593783899429423869404453535514142240725738237814424208303296382145925585774588652778113072312704658578565729252450789704376486755030303476770273209650026296904001464121404548576383
******************
number to be tested or 0 to quit:
1258348708567569132243547867193996784363281
D = -11, dP = 1, P = 1 32768
D = -187, dP = 2, P = 1 4545336381788160000 -3845689020776448000000
j = 1258348708567569132243545594525805890283281
D = -595, dP = 4, P = 1 1908606683491595666107623383040000 8752111455147508300981595950899265536000 483054636550112292687021684688517332992000000 -91399742601830803813322386656934773129216000000
j = 1086925400330382964073177132346989006773405
N[0] = 1258348708567569132243547867193996784363281
a = 1216279372824046067721180339712273625754870
b = 107146466473004570453700290919309043673442
m = 1258348708567569132245765614111687379850544
q = 191857102865472574188504061
P = (141383339, 440369271776032351132821693841488407256480)
P1 = (0, 1)
P2 = (614744102939792574837803939756010891496586, 933339603822832156049176023097197804436211)
N[1] = 191857102865472574188504061
a = 191857102865472572969391004
b = 0
m = 191857102865499431629845250
q = 767428411461997726519381
P = (1782381373, 35592862107491066287643185)
P1 = (0, 1)
P2 = (136132592456267607750070095, 155732348874876656926637504)
N[2] = 767428411461997726519381
a = 767427579541704257876772
b = 0
m = 767428411463744976175172
q = 110783901770629
P = (2066487189, 618541189099874314852572)
P1 = (0, 1)
P2 = (629813497343601678871866, 202173885394664278429567)
N[3] = 110783901770629
a = 110783713479579
b = 0
m = 110783886744834
q = 6154660374713
P = (1504947915, 79263792555867)
P1 = (0, 1)
P2 = (85383958304154, 80332212198595)
N[4] = 6154660374713
a = 6154660374712
b = 0
m = 6154665336208
q = 14915917
P = (2006943890, 1428084754618)
P1 = (0, 1)
P2 = (5298792201095, 2642777147295)
proven prime
number to be tested or 0 to quit:
[/CODE]

Unregistered 2009-12-11 15:12

[QUOTE=cheesehead;198495]Check your assumptions.[/QUOTE]

I guess I/we should check there _is_ a unique 'm' - but I can't bring myself to count all those zeros!!! (yet) [Is this what you mean, cheesehead?] [and anyway, _unless_ I'm missing something obvious(?) it's a stupid/difficult(!) question w/o a unique 'm'...]
J

Unregistered 2009-12-11 15:34

[QUOTE=cheesehead;198495]Check your assumptions.[/QUOTE]

I don't _think_ the 10^m can be split, can it?? [so as to give: pq = (2^m'*5^m''+a)*(5^m'*2^m''+b)]
J

Unregistered 2009-12-11 15:50

Hey again,

thanks a lot for so many replies, did not expect them so quickly :-) Will now look into the problem again. I talked to my prof today and he told me that another student has already sent a solution in.

I also got to know that it cannot be easily brute forced, the key is to think about how these kind of numbers 100000234234000000002342342 can be created.

Thanks again,

Greetings, Chris


All times are UTC. The time now is 01:21.

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.