### How to try to predict the output of Micali-Schnorr Generator (MS-DRBG) knowing the factorization. Part II

See also Part I and Part III of this series

tl;dr In the previous article of the same series we tried to predict the output of Micali-Schnorr Generator (MS-DRBG) knowing the factorization. In this blog post we continue the effort started in part I showing different strategies. If you want to skip all my failures and go directly to the (in my humble opinion) most promising approach you can read directly the

tl;dr In the previous article of the same series we tried to predict the output of Micali-Schnorr Generator (MS-DRBG) knowing the factorization. In this blog post we continue the effort started in part I showing different strategies. If you want to skip all my failures and go directly to the (in my humble opinion) most promising approach you can read directly the

**Solinas Prime and Generalized Mersenne Numbers**section below.
If you actually wonder what is MS-DRBG and why I am trying to do it I'd suggest to go back and read the first article.

What I am

**claiming in this post though is that there is a NSA's backdoor in the ANSI and ISO standards.**__NOT__# Introduction and Failure #1

So let's start from were we actually finished the last post. We focused on an easier version of the problem directly extracted from the original Micali Schnorr paper

from Efficient, perfect random number generators |

**where the known output is up to 3/4 of the RSA computation and secret state is only 1/4 of the RSA computation.**

Assuming we know the factorization of N = p * q we were using Crandall's primes in order to the

*of (s)^e mod(p) and (s)^e mod (q) with s being the secret state at step i-1.*

**least significant bits**
This at first sight looked cool and new to me but I quickly realized, having had some nice reviews and conversations (THANKS!!!), that this result was not too useful....

# Failure #2

This was the phase were I was thinking I was using techniques that are too trivial. All I described in the previous post is, at the end of the day, really trivial math that is known by any Computer Science student and if this problem has been open for so many years it

**MUST**mean that I need to use some fancy stuff if I really want to solve this riddle. Right?
This happened at the same time of ROCA vulnerability that used some lattice sophisticated attack first discovered by Coppersmith. So I tried, as hard as I could. Nothing came out from it....

# Failure #3

I have to admit I was really close to give up at this point, but I have decided to go a bit back on my path and review some of the assumptions I made hoping to find some caveats in my reasoning. If you remember one of the think I quickly dismissed previously was the usage of Mersenne primes (because the lack of the 2^128,2^520 interval, the one interesting for cryptography). This is actually a pity because Mersenne primes have some really beautiful properties. One peculiar one is that if you multiply a number x for 2^k modulo a Mersenne prime, this corresponds to nothing more than a left-circular shift by k bits:

from "Prime Numbers: A Computational Perspective" |

This is really interesting indeed. Why? Let's see.

Taking the same example reported in Prime Numbers: A Computational Perspective

take N = 2^17 -1 = 131071 = 111111111111111112, x = 8977 = 100011000100012 and consider the product 2^5 * x (mod N). This will be the left-circular shift of x by 5 bits, or 25122 = 1100010001000102

Now let's do a couple of observations about this point:

Taking the same example reported in Prime Numbers: A Computational Perspective

take N = 2^17 -1 = 131071 = 111111111111111112, x = 8977 = 100011000100012 and consider the product 2^5 * x (mod N). This will be the left-circular shift of x by 5 bits, or 25122 = 1100010001000102

Now let's do a couple of observations about this point:

- 2^17 * x (mod N) = x indeed 2^17 * 25122 (mod 131071) = 25122
- 2^(17*i) * x (mod N) = x for any integer i > 1

from "Handbook of Elliptic and Hyperelliptic Curve Cryptography" |

Basically reduction modulo a Mersenne prime is really trivial (See above).

### Playing with Mersenne primes and Chinese Remainder Theorem

Now you might ask, why those 2 Mersenne primes's properties are so interesting for Micali-Schnorr Generator (MS-DRBG) ? Before to answer this question we need to open a parenthesis and talk about Chinese Remainder Theorem (CRT from now on). We already mentioned CRT in the previous post where I highlighted that, even I, successfully used CRT in a couple of crypto attacks [0] and [1]. But we never talked about the CRT algorithm per se. The great property of the CRT algorithm is that is reasonably efficient and it is based on the idea of divide a big problem into many little problems. Then, once the little problems are solved, there is a reconstruction phase.

There are different methods for CRT but I would like to focus on one in particular (listed in "An Introduction to Mathematical Cryptography"):

This sequence contains an expansion of the first 1000 entries

As usual N = p * q and we know how to factorize. At step i we will have the MS-DRBG instance looking like (assuming s is the unknown internal state, and z the output of the generator):

And following the CRT method seen above

2^(25) * x (mod 1082401) = x

So finally we have our p = 31 and q = 1082401 and we can apply the CRT trick seen above.

Let spell this down, we just got the

from "An Introduction to Mathematical Cryptography" |

### Mersenne primitive part and cyclotomic polynomial

Now let's us put all the pieces together. We have seen that Mersenne primes have some beautiful properties and an algorithm for CRT. But as we know Mersenne primes are relatively
scarce (specially in some interval as highlighted previously). "Luckily" there are some other numbers (not all primes) that have the same

**"magic"**properties:
Namely

*Mersenne primitive part and cyclotomic polynomial*see below the OEIS sequence (bonus track FWIW some years ago I managed to have my own little sequence listed there) .This sequence contains an expansion of the first 1000 entries

**. But what can we do with those numbers?**

#### Check for plaintext

Let's assume there are 2 distinct Mersenne primes : p=2^7-1 and q=2^14-1**(I know 2^14-1 is not a prime but please trust me and bear with me just a bit more)**.

As usual N = p * q and we know how to factorize. At step i we will have the MS-DRBG instance looking like (assuming s is the unknown internal state, and z the output of the generator):

M = s * 2^14 + z (mod N)

If we know p and q we also know that :
M (mod p)= s * 2^14 (mod p) + z (mod p)

and
M (mod q) = s * 2^14 (mod q) + z (mod q)

Now s is still unknown but if we put all the things we learn so far is clear that s * 2^14 e(mod p) + z (mod p) = s + z (mod p)and s * 2^14 (mod p) + z (mod q) = s + z (mod q) **(do you remember this beautiful property of Mersenne primes?)**.

And following the CRT method seen above

s + zp + p*y = s + zq (mod q)

that is equal to
zp + p*y = zq (mod q)

Now given the fact gcd(p,q) = 1 we can calculate the value of y and eventually the *of R = M (mod p*q).*

**most significant bits**
2^5 * x (mod 31) = x

but also

2^(5*5) * x (mod 31) = 2^(25) * x (mod 31) = x

**but most importantly**

So finally we have our p = 31 and q = 1082401 and we can apply the CRT trick seen above.

Let spell this down, we just got the

**most significant bits****of***M(mod p*q)without knowing the value of s.***The problem is that the missing bits correspond to s itself sigh :(**# The night is darkest just before the dawn

What we got so far is not too useful again. Lets recap:

The effort so far can be summarized by this awesome image from Ange Albertini

I know we are already looking at an easier version of the problem but maybe we can be even less ambitious and lower even more the bar. Well it turns out that for this specific problem this is actually possible :) Indeed (as explained in a comment here) there is a prime-modulus variant that was introduced as an open problem at Micali and Schnorr, "Efficient, Perfect Polynomial Random Number Generators", chapter 4.

But is this really easier?

I was basically stuck until something on the Wikipedia entry for Mersenne primes caught my attention:

Where I heard this Solinas before? Oh of course, Jerry Solinas is the NSA employee that authored the NIST curves and other stuff (*cough* Dual_EC_DRBG ? *cough*)... mmm interesting.

##

So let's give a look to this Solinas prime.

Cool, so we have a modular reduction algorithm that use linear feedback shift register. Interesting. Let's give a look to the original Solinas paper : Generalized Mersenne Numbers. So the example discussed in the paper shows a prime of the form:

then

Hence

- In part I we managed to retrieve
of (s)^e mod(p) and (s)^e mod (q)**least significant bits** - in this post, so far we are able to guess the
**most significant bits****of**M(mod p*q)

The effort so far can be summarized by this awesome image from Ange Albertini

So back to square one :(new comic: Expectations and Realityhttps://t.co/KWwLsjyNLF pic.twitter.com/DkP9gnpsEZ— 👼Ąż杏 (@angealbertini) October 29, 2017

I know we are already looking at an easier version of the problem but maybe we can be even less ambitious and lower even more the bar. Well it turns out that for this specific problem this is actually possible :) Indeed (as explained in a comment here) there is a prime-modulus variant that was introduced as an open problem at Micali and Schnorr, "Efficient, Perfect Polynomial Random Number Generators", chapter 4.

from Efficient, perfect random number generators |

But is this really easier?

I was basically stuck until something on the Wikipedia entry for Mersenne primes caught my attention:

from Wikipedia |

Where I heard this Solinas before? Oh of course, Jerry Solinas is the NSA employee that authored the NIST curves and other stuff (*cough* Dual_EC_DRBG ? *cough*)... mmm interesting.

##
**Solinas Prime and Generalized Mersenne Numbers**

So let's give a look to this Solinas prime.from Wikipedia |

p = 2^3k -2^k +1

The modular reduction obtained the linear feedback shift register is the following

from Generalized Mersenne Numbers |

and here is the numerical example:

from Generalized Mersenne Numbers |

This is indeed really computationally efficient but it is not what we are interested at. When I looked at this the first question that came to my mind was:

**is the algorithm reversible?**Namely is it possible given the result (or part of the result) obtain the original number? Well of course the answer is**NO**. But is it possible for some peculiar number. Well let's take in consideration a special example, where- A5 = A4 = A2 and
- A0 = A3

from Generalized Mersenne Numbers |

from Generalized Mersenne Numbers |

- B0 = -A5

- B1 = A1+A3

**B2 = A4 = A5**

Nice. If we try to map this to our original Micali-Schnorr problem this might be seen as:

- B0 + B1*2^k is the
**output of the generator.**

- B2 is the
**internal state (the one we want to recover)**.

Well it now turns out that for this particular instance of the problem it is possible to recover the internal state given the output. Indeed

**B2 = A4 = A**

**5**

**But we know -A5 = B0 so we can easily recover the state!!!**Now how likely is that we hit this number? Well a rough calculation says that the probability is really low :( and if I am not mistaken is roughly 1/2^3k. What is encouraging though is that we obtained this result from a NSA published paper that used one generalization

from Generalized Mersenne Numbers |

My true hope and

__is that if exists another generalization that gives better result. I am truly convinced the numbers published in the ANSI X9.82 and ISO standards are following the same idea used in Solinas prime and my next goal is to work to prove it.__

**OPEN QUESTION**# Conclusions

In this blog post we again tried to predict the output of Micali-Schnorr Generator (MS-DRBG) knowing the factorization (and possible backdoor it). If any backdoor exists in the ANSI and ISO standards I think there is a possibility that the Solinas primes are used somehow.

My hope is, as always, that my work will inspire someone else to finally find the final solution for this great riddle. Any comment and critique is more than welcome!

My hope is, as always, that my work will inspire someone else to finally find the final solution for this great riddle. Any comment and critique is more than welcome!

# Acknowledgement

I would like to thank Matthew Green to give me such a good puzzle :) and Nadia Heninger.

**That's all folks. For more crypto goodies, follow me on Twitter.**
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