Hey! I'm David, a security engineer at the Blockchain team of Facebook, previously a security consultant for the Cryptography Services of NCC Group. This is my blog about cryptography and security and other related topics that I find interesting.

# Implementation of Boneh and Durfee attack on RSA's low private exponents posted March 2015

I've Implemented a Coppersmith-type attack (using LLL reductions of lattice basis). It was done by Boneh and Durfee and later simplified by Herrmann and May. The program can be found on my github.

The attack allows us to break RSA and the private exponent d. Here's why RSA works (where e is the public exponent, phi is euler's totient function, N is the public modulus):

$ed = 1 \pmod{\varphi(N)}$ $\implies ed = k \cdot \varphi(N) + 1 \text{ over } \mathbb{Z}$ $\implies k \cdot \varphi(N) + 1 = 0 \pmod{e}$ $\implies k \cdot (N + 1 - p - q) + 1 = 0 \pmod{e}$ $\implies 2k \cdot (\frac{N + 1}{2} + \frac{-p -q}{2}) + 1 = 0 \pmod{e}$

The last equation gives us a bivariate polynomial $f(x,y) = 1 + x \cdot (A + y)$. Finding the roots of this polynomial will allow us to easily compute the private exponent d.

The attack works if the private exponent d is too small compared to the modulus: $d < N^{0.292}$.

To use it:

• look at the tests in boneh_durfee.sage and make your own with your own values for the public exponent e and the public modulus N.
• guess how small the private exponent d is and modify delta so you have d < N^delta
• tweak m and t until you find something. You can use Herrmann and May optimized t = tau * m with tau = 1-2*delta. Keep in mind that the bigger they are, the better it is, but the longer it will take. Also we must have 1 <= t <= m.
• you can also decrease X as it might be too high compared to the root of x you are trying to find. This is a last recourse tweak though.

Here is the tweakable part in the code:

# Tweak values here !
delta = 0.26 # so that d < N^delta
m = 3        # x-shifts
t = 1        # y-shifts # we must have 1 <= t <= m
Well done! You've reached the end of my post. Now you can leave me a comment :)

#### ralpo08

There's a mistake on the fourth line of the equations:
\varphi(N) = (p - 1).(q-1) = p.q - p - q + 1 = N - p - q + 1

#### david

good catch! Thanks