I'm trying to understand the ROCA attack on RSA from Matus Nemec et al. but I'm stuck on how they goes from the constraint they have expressed has: $$f(x) = x ∗ M' + (65537^{a'} \mod M') \pmod p$$

To the real polynomial they feed to Coppersmith: $$f(x) = x + (M'^{-1} \mod N) * (65537^{a'} \mod M') \pmod N $$

Specifically how we can go from a polynomial on $\Bbb{Z}/p\Bbb{Z}$ to a polynomial on $\Bbb{Z}/N\Bbb{Z}$ ?

If we try with some real values and we don't use $M'$ but keep $M$: $ M = P_{39}\# \\ a = 1675986788854043070, k = 26617369843\\ \begin{align} p = & k * M + (65537^a\mod M) \\ = & 256311276376047921060658369130455899807 \\ & 39897944295144398331573049960294370089 \end{align} $

$q$ computed in the same way:

$ \begin{align} q = & 143831813798290446194046706419144504095 \\ & 59736324729154061604353910339692872653 \end{align} $

$k$ is obviously a root for

$f(x) = x ∗ M + (65537^{a} \mod M) \pmod p$

but, as far as I understand it's not for

$f(x) = x + (M^{-1} \mod N) * (65537^{a} \mod M) \pmod N $

With: $ \begin{align} M^{-1} \mod N = & 352742121330634029642965446346501378229514125041483823729 \\ & 9285302994689783984115678100687263685016509742529377 \\ & 76329363423355363257168553996151098868709524 \end{align} $

Where am I wrong on this?

  • $\begingroup$ Take a look at theorems 5 and 6 in Alexander May's PhD thesis. You need the conditions (1) $p^k$ divides $f(x)$ and (2) $|f(x)| < p^k$ to get that $x$ is a root. Condition (2) you obtain by viewing $f(x)$ as a scalar product of a vector containing the coefficients of the polynomial $f$ and a vector containing the powers of $x$. $\endgroup$ – j.p. May 28 '18 at 5:58
  • $\begingroup$ Sorry, my $k$ doesn't have to do with your $k$. $\endgroup$ – j.p. May 28 '18 at 19:37

You are correct—you do want to find a small root of [*] $$ f(x) = M \cdot x + (65537^a \bmod M) \bmod N $$ modulo a divisor $p$ of $N$. In other words, you want to find a divisor of $N$ in the residue class $M \cdot x + (65537^a \bmod M)$. This is explicitly accomplished by the Coppersmith/Howgrave-Graham method, as long as the appropriate restrictions are respected.

However, Coppersmith's method to find such roots traditionally expects a monic polynomial as input. So, to make it monic we simply divide it by the coefficient associated with $x$, that is, $M$: $$ g(x) = \frac{1}{M}f(x) = x + \frac{65537^a \bmod M}{M} \bmod N\,. $$ The roots of $g(x)$ are the same as $f(x)$; to see this decompose the polynomial into its factorization: $$ f(x) = (x - \alpha_0)(x - \alpha_1)\dots(x - \alpha_n)\,, $$ $$ g(x) = \frac{1}{M}(x - \alpha_0)(x - \alpha_1)\dots(x - \alpha_n)\,, $$ and notice that both have the same roots $\alpha_i$. The Coppersmith method finds the appropriate root modulo a factor of $N$ once you guess the correct $a$.

[*]: The ROCA paper uses alternative $M'$ there, but this is irrelevant here.

  • $\begingroup$ I thought the constraint was $f(x) = M \cdot x + (65537^a \mod M) \mod p$ and not $f(x) = M \cdot x + (65537^a \mod M) \mod N$. $\endgroup$ – needle Aug 3 '18 at 14:15
  • $\begingroup$ It is more convenient to use the Alexander May (§3.2) formulation of Coppersmith's method, which explicitly finds a small root of $f(x)$ modulo an unknown factor of $N$, i.e., $p$. $\endgroup$ – Samuel Neves Aug 7 '18 at 9:55

One of the important part of the attack is to find $f$ such as $f(x) \equiv 0\bmod p$.

First, you reduce $f(x)= x + (M^{-1} \bmod N) * (65537^{a} \bmod M)$ modulo $N$, to get a smaller coefficient (we can do that because $p$ divides $N$), and then we can check if we still have $f(k) \equiv 0 \bmod p$:

\begin{array}{rcll} f(k) & \equiv & k + (M^{-1} \bmod N)(65537^{a} \bmod M) & \mod p \\ & \equiv & k + (M^{-1} \bmod N)(-kM) & \mod p \\ & \equiv & k - k & \mod p \\ & \equiv & 0 & \mod p. \end{array}

On the second line because $p=kM+(65537^a\bmod M)$ so $(65537^a\bmod M) \equiv -kM \bmod p$, and the third line because $M(M^{-1} \bmod N) \equiv 1 \bmod p$.


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