0
$\begingroup$

I read in literature that Rabin Cryptosystem can be broken using chosen-ciphertext attack. It is described that after chosen ciphertext is decrypted attacker can factorize public key $n$ by using square root with probability of $1/2$. But in article it is not described how this factorization is done.

If somebody can give some example I would be grateful.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

I'll assume this is not homework.

It's actually quite simple:

  • Pick a random value $r$

  • Compute $s = r^2 \bmod n$, and submit $s$ to the Rabin decryptor

  • Since $s$ is a Quadratic Residue, the Rabin decryptor will return some value $t = \sqrt{s} \bmod n$.

  • Now, $s$ has four square roots (assuming $n$ has two prime factors and you didn't happen to pick an $r$ that's not relatively prime to $n$); if you have $t = r$ or $t = n-r$, it didn't work. If it's one of the other two possible values, then we have $r^2 = t^2 \bmod n$, that is, $(r+t)(r-t) = kn$, for some integer $k$; however neither $r+t$ nor $r-t$ are multiples of $n$, hence $\gcd(r+t, n)$ is a nontrivial factor of $n$ (and $\gcd(r-t, n)$ is another nontrivial factor).

Because we selected $r$ at random, Rabin has no way of knowing which of the four possibilities we picked; it returns one of them, and so the probability that it happens to return one that gives us the factorization is $2/4 = 0.5$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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