mersenneforum.org math gcd problem need help
 Register FAQ Search Today's Posts Mark Forums Read

 2017-09-23, 10:59 #45 crack11   Sep 2017 23·3 Posts dear sir if you don t mind can you pm your skype id than i give you more exmpel..
 2017-09-23, 13:26 #46 Dr Sardonicus     Feb 2017 Nowhere 2×13×137 Posts Curiously, the OP's original question was posted to a Wolfram Stackexchange chat room the day before it was posted to this august Forum. The user zhk had identified the trouble as an overflow problem. I also found an old thread in this Forum with exactly the same kind of calculation as in this one. Major difference: When asked, the OP in that thread described precisely what he was trying to do and why: It's possible to calculate an unknown RSA modulus?
 2017-09-23, 13:43 #47 crack11   Sep 2017 23·3 Posts yes DR sir my calculating string same like this:http://www.mersenneforum.org/showthread.php?t=13524
2017-09-23, 14:36   #48
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

2·3·5·47 Posts

Quote:
 Originally Posted by axn It looks to me, from the numbers posted above, that they know the plaintext/ciphertext pairs, but doesn't know the N, and the gcd is supposed to give the N? Weird.
Yes, exactly that!
Code:
m1^e==c1 mod N  [1]
m2^e==c2 mod N  [2]
from here: N | gcd(Z1,Z2)=gcd(m1^e-c1,m2^e-c2), because N|Z1,Z2, but it is NOT TRUE, that gcd(Z1,Z2)=N. But since for random numbers gcd is "small", the gcd(Z1,Z2) will be a small multiple of N, hence you (not me) can "easily" factor that, and using [1] or [2] we can easily choose the true value of N (in almost every cases) as a divisor of gcd(Z1,Z2). The first hard step is to get gcd(Z1,Z2).

And why would anybody solve this problem where the original message is KNOWN , these are m1 and m2. Or there are other c values, where m is unknown.

 2017-09-23, 15:00 #49 Dr Sardonicus     Feb 2017 Nowhere 2×13×137 Posts We've repeatedly told the OP "The numbers are too big" Repeatedly ignored.
 2017-09-23, 16:25 #50 chris2be8     Sep 2009 23·239 Posts Could it be done by calculating everything mod some suitable prime larger than N, eg 2^521-1 (assuming N is RSA512)? Or mod several smaller primes and applying the Chinese Remainder Theorem to recover N? Chris
 2017-09-25, 13:15 #51 crack11   Sep 2017 23·3 Posts Dear all. my string same like this N modul recover..
2017-09-25, 13:20   #52
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by crack11 Dear all. my string same like this N modul recover..
there are very basic results you could get if you tried. but since you don't seem to know about PARI/GP etc. odds are you don't know what my comment means.

 2020-08-13, 04:15 #53 crack11   Sep 2017 23×3 Posts no body try to solve this issu overflow?
2020-08-13, 06:50   #54
paulunderwood

Sep 2002
Database er0rr

22·3·7·41 Posts

Quote:
 Originally Posted by crack11 no body try to solve this issu overflow?
Arbitrary precision arithmetic is a solved problem.

 2020-08-13, 10:45 #55 crack11   Sep 2017 23×3 Posts dear sir, your send link not solve the probleam its all ready solve like old exp-65537 5 digt but new exp-12 digit can it possibel with exp calculating start?

 Similar Threads Thread Thread Starter Forum Replies Last Post dabler Miscellaneous Math 1 2018-07-28 14:03 ET_ Operazione Doppi Mersennes 4 2012-09-20 19:33 jasong Homework Help 10 2012-04-21 01:09 jinydu Puzzles 4 2003-12-13 06:00 daxm Miscellaneous Math 5 2003-07-20 19:32

All times are UTC. The time now is 03:48.

Fri Oct 23 03:48:34 UTC 2020 up 43 days, 59 mins, 0 users, load averages: 2.19, 1.67, 1.57