## Multi-prime RSA

June 26, 2020 *in Factorization, 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