Rabin cryptosystem

The Rabin cryptosystem is an asymmetric cryptographic technique, whose security, like that of RSA, is related to the difficulty of integer factorization. However the Rabin cryptosystem has the advantage that it has been mathematically proven to be computationally secure against a chosen-plaintext attack as long as the attacker cannot efficiently factor integers, while there is no such proof known for RSA.:145 It has the disadvantage that each output of the Rabin function can be generated by any of four possible inputs; if each output is a ciphertext, extra complexity is required on decryption to identify which of the four possible inputs was the true plaintext.

History

The algorithm was published in January 1979 by Michael O. Rabin. The Rabin cryptosystem was the first asymmetric cryptosystem where recovering the plaintext from the ciphertext could be proven to be as hard as factoring.

Encryption Algorithm

Like all asymmetric cryptosystems, the Rabin system uses a key pair: a public key for encryption and a private key for decryption. The public key is published for anyone to use, while the private key remains known only to the recipient of the message.

Key generation

The keys for the Rabin cryptosystem are generated as follows:

1. Choose two large distinct prime numbers $p$ and $q$ such that $p\equiv 3{\bmod {4}}$ and $q\equiv 3{\bmod {4}}$ .
2. Compute $n=pq$ .

Then $n$ is the public key and the pair $(p,q)$ is the private key.

Encryption

A message $M$ can be encrypted by first converting it to a number $m using a reversible mapping, then computing $c=m^{2}{\bmod {n}}$ . The ciphertext is $c$ .

Decryption

The message $m$ can be recovered from the ciphertext $c$ by taking its square root modulo $n$ as follows.

1. Compute the square root of $c$ modulo $p$ and $q$ using these formulas:
{\begin{aligned}m_{p}&=c^{{\frac {1}{4}}(p+1)}{\bmod {p}}\\m_{q}&=c^{{\frac {1}{4}}(q+1)}{\bmod {q}}\end{aligned}} 2. Use the extended Euclidean algorithm to find $y_{p}$ and $y_{q}$ such that $y_{p}\cdot p+y_{q}\cdot q=1$ .
3. Use the Chinese remainder theorem to find the four square roots of $c$ modulo $n$ :
{\begin{aligned}r_{1}&=\left(y_{p}\cdot p\cdot m_{q}+y_{q}\cdot q\cdot m_{p}\right){\bmod {n}}\\r_{2}&=n-r_{1}\\r_{3}&=\left(y_{p}\cdot p\cdot m_{q}-y_{q}\cdot q\cdot m_{p}\right){\bmod {n}}\\r_{4}&=n-r_{3}\end{aligned}} One of these four values is the original plaintext $m$ , although which of the four is the correct one cannot be determined without additional information.

Computing square roots

We can show that the formulas in step 1 above actually produce the square roots of $c$ as follows. For the first formula, we want to prove that $m_{p}^{2}\equiv c{\bmod {p}}$ . Since $p\equiv 3{\bmod {4}},$ the exponent ${\textstyle {\frac {1}{4}}(p+1)}$ is an integer. The proof is trivial if $c\equiv 0{\bmod {p}}$ , so we may assume that $p$ does not divide $c$ . Note that $c\equiv m^{2}{\bmod {pq}}$ implies that $c\equiv m^{2}{\bmod {p}}$ , so c is a quadratic residue modulo $p$ . Then

$m_{p}^{2}\equiv c^{{\frac {1}{2}}(p+1)}\equiv c\cdot c^{{\frac {1}{2}}(p-1)}\equiv c\cdot 1\mod p$ The last step is justified by Euler's criterion.

Example

As an example, take $p=7$ and $q=11$ , then $n=77$ . Take $m=20$ as our plaintext. The ciphertext is thus $c=m^{2}{\bmod {n}}=400{\bmod {77}}=15$ .

Decryption proceeds as follows:

1. Compute $m_{p}=c^{{\frac {1}{4}}(p+1)}{\bmod {p}}=15^{2}{\bmod {7}}=1$ and $m_{q}=c^{{\frac {1}{4}}(q+1)}{\bmod {q}}=15^{3}{\bmod {11}}=9$ .
2. Use the extended Euclidean algorithm to compute $y_{p}=-3$ and $y_{q}=2$ . We can confirm that $y_{p}\cdot p+y_{q}\cdot q=(-3\cdot 7)+(2\cdot 11)=1$ .
3. Compute the four plaintext candidates:
{\begin{aligned}r_{1}&=(-3\cdot 7\cdot 9+2\cdot 11\cdot 1){\bmod {77}}=64\\r_{2}&=77-64=13\\r_{3}&=(-3\cdot 7\cdot 9-2\cdot 11\cdot 1){\bmod {77}}=\mathbf {20} \\r_{4}&=77-20=57\end{aligned}} and we see that $r_{3}$ is the desired plaintext. Note that all four candidates are square roots of 15 mod 77. That is, for each candidate, $r_{i}^{2}{\bmod {77}}=15$ , so each $r_{i}$ encrypts to the same value, 15.

Digital Signature Algorithm

The Rabin cryptosystem can be used to create and verify digital signatures. Creating a signature requires the private key $(p,q)$ . Verifying a signature requires the public key $n$ .

Signing

A message $m can be signed with a private key $(p,q)$ as follows.

1. Generate a random value $u$ .
2. Use a cryptographic hash function $H$ to compute $c=H(m|u)$ , where the bar denotes concatenation. $c$ should be an integer less than $n$ .
3. Treat $c$ as a Rabin-encrypted value and attempt to decrypt it, using the private key $(p,q)$ . This will produce the usual four results, $r_{1},r_{2},r_{3},r_{4}$ .
4. One might expect that encrypting each $r_{i}$ would produce $c$ . However, this will be true only if $c$ happens to be a quadratic residue modulo $p$ and $q$ . To determine if this is the case, encrypt the first decryption result $r_{1}$ . If it does not encrypt to $c$ , repeat this algorithm with a new random $u$ . The expected number of times this algorithm needs to be repeated before finding a suitable $c$ is 4.
5. Having found an $r_{1}$ which encrypts to $c$ , the signature is $(r_{1},u)$ .

Verifying a signature

A signature $(r,u)$ for a message $m$ can be verified using the public key $n$ as follows.

1. Compute $c=H(m|u)$ .
2. Encrypt $r$ using the public key $n$ .
3. The signature is valid if and only if the encryption of $r$ equals $c$ .

Evaluation of the algorithm

Effectiveness

Decrypting produces three false results in addition to the correct one, so that the correct result must be guessed. This is the major disadvantage of the Rabin cryptosystem and one of the factors which have prevented it from finding widespread practical use.

If the plaintext is intended to represent a text message, guessing is not difficult; however, if the plaintext is intended to represent a numerical value, this issue becomes a problem that must be resolved by some kind of disambiguation scheme. It is possible to choose plaintexts with special structures, or to add padding, to eliminate this problem. A way of removing the ambiguity of inversion was suggested by Blum and Williams: the two primes used are restricted to primes congruent to 3 modulo 4 and the domain of the squaring is restricted to the set of quadratic residues. These restrictions make the squaring function into a trapdoor permutation, eliminating the ambiguity.

Efficiency

For encryption, a square modulo n must be calculated. This is more efficient than RSA, which requires the calculation of at least a cube.

For decryption, the Chinese remainder theorem is applied, along with two modular exponentiations. Here the efficiency is comparable to RSA.

Disambiguation introduces additional computational costs, and is what has prevented the Rabin cryptosystem from finding widespread practical use.[citation needed]

Security

It has been proven that any algorithm which decrypts a Rabin-encrypted value can be used to factor the modulus $n$ . Thus, Rabin decryption is at least as hard as the integer factorization problem, something that has not been proven for RSA. It is generally believed that there is no polynomial-time algorithm for factoring, which implies that there is no efficient algorithm for decrypting a Rabin-encrypted value without the private key $(p,q)$ .

The Rabin cryptosystem does not provide indistinguishability against chosen plaintext attacks since the process of encryption is deterministic. An adversary, given a ciphertext and a candidate message, can easily determine whether or not the ciphertext encodes the candidate message (by simply checking whether encrypting the candidate message yields the given ciphertext).

The Rabin cryptosystem is insecure against a chosen ciphertext attack (even when challenge messages are chosen uniformly at random from the message space).:150 By adding redundancies, for example, the repetition of the last 64 bits, the system can be made to produce a single root. This thwarts this specific chosen-ciphertext attack, since the decryption algorithm then only produces the root that the attacker already knows. If this technique is applied, the proof of the equivalence with the factorization problem fails, so it is uncertain as of 2004 if this variant is secure. The Handbook of Applied Cryptography by Menezes, Oorschot and Vanstone considers this equivalence probable, however, as long as the finding of the roots remains a two-part process (1. roots ${\bmod {p}}$ and ${\bmod {q}}$ and 2. application of the Chinese remainder theorem).