Saduff Posted August 15, 2011 Share Posted August 15, 2011 Since my last KGM was too simple, then here's another one. This one should be harder.KeygenMe16.zip Link to comment Share on other sites More sharing options...
MA1201 Posted August 15, 2011 Share Posted August 15, 2011 Name : k‰ÔØÿv|EüPÔ£w«UX£‹:"§‰¦ùX Serial : C29FE4713F31B48A02C3E58D803745EE4A9BFE01976B97707975ED312F751DF2F4E4CC307EEC571722F40454A1595E24B699518878163C256B16AEDF11A65B82 i know it's lame, just for lulz hope it works.. Link to comment Share on other sites More sharing options...
Saduff Posted August 15, 2011 Author Share Posted August 15, 2011 Unfortunately, doesn't work. Name should be:k‰ŌŲ˙v|EüPŌ£w«UX£‹:"§‰¦łXName you gave:k‰ÔØÿv|EüPÔ£w«UX£‹:"§‰¦ùX Link to comment Share on other sites More sharing options...
MA1201 Posted August 15, 2011 Share Posted August 15, 2011 plz check again, the name you provided doesn't work, the name & serial combination i posted works Link to comment Share on other sites More sharing options...
Saduff Posted August 15, 2011 Author Share Posted August 15, 2011 (edited) Hmm, well, on my machine (XP SP3 x86), the name I provided works, but not the one you provided. Edited August 15, 2011 by Saduff Link to comment Share on other sites More sharing options...
KKR_WE_RULE Posted August 15, 2011 Share Posted August 15, 2011 The serial provided by Ma1201 works but the you(Saduff) provided aint working Link to comment Share on other sites More sharing options...
HMX0101 Posted August 17, 2011 Share Posted August 17, 2011 I know what's going on here, but I'm not sure if the other value from public-key pair (since you're using it backwards) can be retrieved. Link to comment Share on other sites More sharing options...
qpt^J Posted August 17, 2011 Share Posted August 17, 2011 (edited) we need to get G from equation U = G^L (mod N^2) whereU = 37F9C8007A2E278F9ED43223657938A693752710D030A71A1FA7D2380FE7BE4CB6E3DFFE13ADB30967E4A7443046E6ECCB75B11A3D932E7F9AA44E02FCA166D7L = 203CDDC2D7EC7FE5CB7352EC3C3DD9616160E54E3B2A7EE3FB039511466D6CD0N^2 = 40F45E9696A53EFB3F7E5A77111E1702826D8734C5E69226DEAD774BE70D7F974F2A0F8BC46107269F7193B4093BBB2ACF0ACA5D2FA08AE1BE29E5BFB4816329if i didnt done any mistake anywhere..is there any known algo to compute this? :Sand i guess homomorphism won't help here too much. and it's understandable why you've used decryption, instead of encryption. but even if you use encryption there, i think solving 512 bit DL problem wont be that easy even for such field Z_N^2EDIT:The problem is:U = 1 / [L(g^l mod n^2)]L(x) = (x-1)/nAs I said, I think its not possible to get G.. the symbol "U" you wrote there is "myu" from greek alphabet i thinkand it's possible to compute the second algo you wrote there.. the only problem is so-called DCRA/>http://en.wikipedia.org/wiki/Decisional_composite_residuosity_assumption Edited August 17, 2011 by qpt^J Link to comment Share on other sites More sharing options...
HMX0101 Posted August 17, 2011 Share Posted August 17, 2011 The problem is:U = 1 / [L(g^l mod n^2)]L(x) = (x-1)/nAs I said, I think its not possible to get G.. Link to comment Share on other sites More sharing options...
Saduff Posted August 17, 2011 Author Share Posted August 17, 2011 Damn, I was afraid it might not be possible to get G. My attempts to get it failed, but I was hoping it's just me and the KGM was already ready, so I published it anyway. Well, in that case, should I publish G, so you could gen it? Link to comment Share on other sites More sharing options...
qpt^J Posted August 18, 2011 Share Posted August 18, 2011 i think theres no fun from that, and maybe some ppl are more smarter here and know how to defeat it, so i guess this cme will not be solved for a quite while if not forever Link to comment Share on other sites More sharing options...
_sb_ Posted August 18, 2011 Share Posted August 18, 2011 The Ideal would be to factor U, That would take a weak or more, But If one can factor N^2 then I think it would be easier to solve this. Link to comment Share on other sites More sharing options...
Saduff Posted August 18, 2011 Author Share Posted August 18, 2011 If numbers are too big, I could recompile with smaller numbers if necessary... Link to comment Share on other sites More sharing options...
_sb_ Posted August 18, 2011 Share Posted August 18, 2011 I will not be near a proper not-browser only computer for a while, I just thought that you had given some thought when you compiled this one and that there would actually be weaknesses on your routine, unfortunantly that was not true it seems. Link to comment Share on other sites More sharing options...
HMX0101 Posted August 19, 2011 Share Posted August 19, 2011 You don't need to factor N^2, since N=256bits and N = p*q, N^2 = (p^2) * (q^2).Reducing numbers won't do any difference.. Link to comment Share on other sites More sharing options...
ChOoKi Posted November 3, 2011 Share Posted November 3, 2011 Damn, I was afraid it might not be possible to get G. My attempts to get it failed, but I was hoping it's just me and the KGM was already ready, so I published it anyway. For what it's worth sol.zip 1 Link to comment Share on other sites More sharing options...
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