lists.openwall.net   lists  /  announce  owl-users  owl-dev  john-users  john-dev  passwdqc-users  yescrypt  popa3d-users  /  oss-security  kernel-hardening  musl  sabotage  tlsify  passwords  /  crypt-dev  xvendor  /  Bugtraq  Full-Disclosure  linux-kernel  linux-netdev  linux-ext4  linux-hardening  linux-cve-announce  PHC 
Open Source and information security mailing list archives
 
Hash Suite for Android: free password hash cracker in your pocket
[<prev] [next>] [thread-next>] [day] [month] [year] [list]
Date: Thu, 26 Apr 2007 15:04:39 +0400
From: Eugene Chukhlomin <chukh29ru@...oline.su>
To: Full-Disclosure <full-disclosure@...ts.grok.org.uk>
Subject: Re: Rapid integer factorization = end of RSA?

>#v+
>gap> p;
>163473364580925384844313388386509085984178367003309231218111085238933310010450\
>8151212118167511579
>gap> q;
>190087128166482211312685157393541397547189678996851549366663853908802710380210\
>4498957191261465571
>gap> n := p * q;
>310741824049004372135075003588856793003734602284272754572016194882320644051808\
>150455634682967172328678243791627283803341547107310850191954852900733772482278\
>3525742386454014691736602477652346609
>gap> (p * (n - q)) mod n;
>0
>gap> 
>#v-
>
>What is it supposed to proove?

  

My gypothesis: if exists subsets(A1...An) and (B1...Bn) which satisfies 
equality: A1*B1 +...An*Bn =  N = p*q, then exists some of them, which 
satisfies equality A1*(-B1)+...An*(-Bn)=p*q*(q-1)

_______________________________________________
Full-Disclosure - We believe in it.
Charter: http://lists.grok.org.uk/full-disclosure-charter.html
Hosted and sponsored by Secunia - http://secunia.com/

Powered by blists - more mailing lists

Powered by Openwall GNU/*/Linux Powered by OpenVZ