Since my last KGM was too simple, then here's another one. This one should be harder.

KeygenMe16.zip

Name : k‰ÔØÿv|EüPÔ£w«UX£‹:"§‰¦ùX

Serial : C29FE4713F31B48A02C3E58D803745EE4A9BFE01976B97707975ED312F751DF2F4E4CC307EEC571722F40454A1595E24B699518878163C256B16AEDF11A65B82

i know it's lame, just for lulz

hope it works..

Unfortunately, doesn't work. Name should be:

k‰ŌŲ˙v|EüPŌ£w«UX£‹:"§‰¦łX

Name you gave:

k‰ÔØÿv|EüPÔ£w«UX£‹:"§‰¦ùX

plz check again, the name you provided doesn't work, the name & serial combination i posted works

Hmm, well, on my machine (XP SP3 x86), the name I provided works, but not the one you provided.

Edited August 15, 201114 yr by Saduff

The serial provided by Ma1201 works but the you(Saduff) provided aint working

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.

we need to get G from equation U = G^L (mod N^2) where

U = 37F9C8007A2E278F9ED43223657938A693752710D030A71A1FA7D2380FE7BE4CB6E3DFFE13ADB30967E4A7443046E6ECCB75B11A3D932E7F9AA44E02FCA166D7

L = 203CDDC2D7EC7FE5CB7352EC3C3DD9616160E54E3B2A7EE3FB039511466D6CD0

N^2 = 40F45E9696A53EFB3F7E5A77111E1702826D8734C5E69226DEAD774BE70D7F974F2A0F8BC46107269F7193B4093BBB2ACF0ACA5D2FA08AE1BE29E5BFB4816329

if i didnt done any mistake anywhere..

is there any known algo to compute this? :S

and 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^2

EDIT:

The problem is:

U = 1 / [L(g^l mod n^2)]

L(x) = (x-1)/n

As I said, I think its not possible to get G..

the symbol "U" you wrote there is "myu" from greek alphabet i think

and 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, 201114 yr by qpt^J

The problem is:

U = 1 / [L(g^l mod n^2)]

L(x) = (x-1)/n

As I said, I think its not possible to get G..

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?

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

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.

If numbers are too big, I could recompile with smaller numbers if necessary...

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.

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..

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