Marrying client-side Windows-based CryptEncrypt and server-side,Linux-based Crypt::OpenSSL::RSA

Time flies and it does so very quickly. The story I am about to tell you is 8 years old, but it does feel like I wrote it yesterday.

In 2014 a client asked me to develop a never-seen-before prototype of a new type of an endpoint agent that would be code-minimal, position-independent, 32- and 64- architecture-aware and talk to a backend using strong encryption. Yes, kinda EDR or C2-like agent and we had discussions about using it for both blue and red team engagements, if it worked.

Anyone who tried to make Windows crypto primitives talk to (typically Linux-based) server-side crypto primitives knows that it is an awful coding experience. After googling around, and trying different things I eventually developed the prototype. I can’t share the code for obvious reasons, but I can at least describe what it did.

On a client side, I had a routine that would talk to the socket (not proxy aware at that time) and follow a basic data protocol exchanging encrypted data blobs with my server. The data was encrypted with a public key that only server could decrypt. Nothing really ground breaking.

What was annoyingly, frustratingly hard to develop was the actual decryption part. The server part was using Crypt::OpenSSL::RSA (yes, perl!) primitive, and I couldn’t force it to decrypt the CryptEncrypted message I was sending.

After many hours of debugging and googling around I eventually figured it out. After I used CryptEncrypt I just had to reverse the data blob delivered by the function: byte, by byte. Yup, it was that simple.

Multi-prime RSA

Today for a change I will talk about a topic that is completely abstract to me. I got interested in it as a side-effect of a project I was working on and here’s a code that some may find useful.

Researching RSA crypto internals you will surely come across a lot of information and practical examples talking about calculation of a private key / decryption if you can only factor ‘n = p & q’. For example this great example.

I was curious how to do it not only for p & q, but also for multi-prime RSA setups, that is – these where ‘n = r[1] x r[2] x … x r[n]’. There is not that much published about it online and the best example I could find was this article on StackExchange.

While I don’t read mathematical symbols very well, I decided to implement it in python to learn something new. In order to keep it flexible, the whole code works based on the content of the ‘r’ array (or list really), that specifies all multipliers of ‘n’. The Modular multiplicative inverse function is stolen from this article on StackOverflow.

Here’s the code:

import math
def egcd(a, b):
   if a == 0:
      return (b, 0, 1)
   else:
      g, y, x = egcd(b % a, a)
   return (g, x - (b // a) * y, y)

def cmi(a, m):
   g, x, y = egcd(a, m)
   if g != 1:
      return None
   else:
      return x % m

r = [931164518537359, 944727352543879, 982273258722607]
y = 529481440313141057262802385309623737292746309
lr = len(r)
e = 5
N = 1
d = [None] * lr
t = [None] * lr
x = [None] * lr
for i in range(lr):
   N = N * r[i]
   d[i] = cmi (e, r[i]-1)
   print ("r[", i , "] = ", r[i])
   print ("N = ", N)

print ()

for i in range(lr):
   print ("d[", i , "] = ", d[i])

print ()

m=r[0]
for i in range(1,lr):
   t[i] = cmi (m, r[i])
   m = m * r[i]
   print ("t[", i , "] = ", t[i])

print ()

for i in range(lr):
   x[i] = pow(y, d[i] ,r[i]) % r[i]
   print ("x[", i , "] = ", x[i])

print ()

X = x[0]
m = r[0]
for i in range(1,lr):
   X = X + m * ( ( (x[i] - (X % r[i]) ) * t[i] ) % r[i] )
   m = m * r[i]
   print ("x = ", X)

Here are the numbers from the StackExchange post:

e=5
r1=931164518537359 r2=944727352543879 r3=982273258722607
N=864102436520313334659779717201860718296307527
d1=558698711122415 d2=566836411526327 d3=785818606978085
t2=360227672914825 t3=882117903741868
y=529481440313141057262802385309623737292746309
x1=436496882968258 x2=903092574358267 x3=806961802724
x=710532117316769399313215266414 (when i=2)
x=111222333444555666777888999000000000000000042

And here is the output of the script:

r[ 0 ] = 931164518537359
r[ 1 ] = 944727352543879
r[ 2 ] = 982273258722607
N = 864102436520313334659779717201860718296307527

d[ 0 ] = 558698711122415
d[ 1 ] = 566836411526327
d[ 2 ] = 785818606978085

t[ 1 ] = 360227672914825
t[ 2 ] = 882117903741868

x[ 0 ] = 436496882968258
x[ 1 ] = 903092574358267
x[ 2 ] = 806961802724

x = 710532117316769399313215266414
x = 111222333444555666777888999000000000000000042

And what about the article I mentioned earlier?

e = 113
p = 338924256021210389725168429375903627261
q = 338924256021210389725168429375903627349
ct = 102692755691755898230412269602025019920938225158332080093559205660414585058354

For which the expected result is:

t :535645912235879621902477379288244888293287927881054157873533

If I plug it in to the code I pasted above, I will get the following:

r = [338924256021210389725168429375903627261, 338924256021210389725168429375903627349]
y = 102692755691755898230412269602025019920938225158332080093559205660414585058354
lr = len(r)
e = 113

And the output is:

r[ 0 ] = 338924256021210389725168429375903627261
r[ 1 ] = 338924256021210389725168429375903627349
N = 114869651319530967114595389434126892905129957446815070167640244711056341561089
d[ 0 ] = 155965144363742834209812020597760961217
d[ 1 ] = 329926266923302149289986966649109725737
t[ 1 ] = 234936132014702656514037206726478650776
x[ 0 ] = 22808800254162271774126722746602450507
x[ 1 ] = 22808800254162132696325593613158654299
x = 535645912235879621902477379288244888293287927881054157873533