If the GCD(r, p-1) is small and the value k is used to sign a message using ElGamal is also small. Then the secret value of x can be determined.

Why is this true? How would one retrieve x?

  • 1
    $\begingroup$ Is this a homework task, or from where did you get this assertion? $\endgroup$ Nov 27 '11 at 13:42
  • $\begingroup$ From an old exam question. Studying for finals $\endgroup$
    – Bobby S
    Nov 28 '11 at 19:10

$\newcommand\gcd{\operatorname{gcd}}$Let's have a look at the signature equation:

$$ s = (H(m) - x·r)·k^{-1} \mod (p-1), $$ $$ s·k = H(m) - x·r \mod (p-1), $$ and thus $$ H(m) - s·k = x · r \mod (p-1).$$

$d = \gcd(r, p-1)$ means we find (efficiently, given $r$ and $p-1$, using the extended euclidean algorithm) a $z$ such that $z·r = d \mod (p-1)$ ... this can be regarded as an "almost-inverse" for $r$ if $d$ is small.

If we multiply the equation above by $z$, we get $$ z·(H(m) - s·k) = x·d \mod (p-1)$$

$s$ and $r$ are the signature, $m$ and $y = g^x$ are also known to the attacker, $d$ is a known small divisor of the modulus, and $k$ is also "small", which means that we can brute-force over all possible values of $k$ and for each one see if there is a solution for $x$.

The moral of the story: Use a random $k$, which will most likely not be small.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.