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There is large body of literature focusing on the cryptanalysis of RSA. An excellent source of information is Boneh's survey [19]. In this section, we mention a few mathematical issues related to the security of RSA. Our intent is to illustrate some of the issues that can lead to attacks on flawed implementations of RSA. First, we mention a generic attack that applies to public key systems (and hash functions in some situations) but not to symmetric ciphers. We state the
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PUBLIC KEY SYSTEMS
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attack in the context of RSA, but it applies to other public key cryptosystems
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Suppose that Alice encrypts the secret message
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using Bob s public key ( e , N ) ,that is, she computes C = M e (niod N ) , and she sends C to Bob. Suppose Trudy intercepts C. Since Bob s public key is public, Trudy can try to guess messages M and for each guess compute the putative ciphertext C = (M ) (mod N ) . If Trudy ever finds a message M for which C = C, then Trudy knows that M = M and she has broken the RSA encryption. This method of attacking public key encryption is known as a forward search. It does not apply to symmetric ciphers, since Alice and Bob s shared symmetric key would not be available to Trudy, so she could not try to encrypt likely messages. Obviously, for both synirnetric key and public key cryptography, the size of the key space must he large enough to prevent a brute-force exhaustive search. However, the forward search attack shows that for public key encryption, the size of the plaintext space must be sufficiently large that an attacker cannot simply try to encrypt all likely messages. In practice, it is easy to prevent the forward search attack by padding messages with a sufficient number of raridorn bits, thereby increasing the size of the plaint,ext space. RSA is also susceptible to a chosen ciphertext attack in the following sense. Suppose that Alice will decrypt an innocent-looking ciphertext of Trudy s choosing and return the result to Trudy. Then Trudy can recover the plaintext for any ciphertext that was encrypted with Alice s public key; see Problem 1 3 . An interesting mathematical fact regarding RSA is that the factors p arid q of N a.re easily obtained if we know q5(N). To see why this is so, suppose that 4 ( N )is known, where 1v = p q with r). < p . Then
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l ) ( q - 1) = p q
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d ( N )+ 1.
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which implies
q = J(p
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From (6.8) and (6.9), we can easily compute
6.5 RSA
and we have factored N . In addition, if 4 ( N ) is known, the private key d is easily recovered by using the Euclidean Algorithm (see Section A-2 in the Appendix), since the encryption exponent e is public knowledge. Since d is the multiplicative inverse of e modulo 4 ( N ) ,the process to determine d is precisely the same as that used in the construction of the key pair. Clearly, the numbers p , q and d must be large, so as to prevent a bruteforce attack, but what other properties should they (and e) have When constructing the modulus N , the prime numbers p and q need to be chosen carefully. A s t r o n g prime p is a prime number such that p - 1 has a large prime factor r , p 1 has a large prime factor, and T - 1 has a large prime factor. Strong primes p and q should always be used in any implementation of RSA to thwart the factorization of N through the use of Pollard s p - 1 Algorithm [all. If the modulus N is misused, then RSA is easily compromised. Suppose that ( e j , N ) are the RSA public keys of j parties. That is, each user has the same modulus N , but a different public encryption exponents e j (and, therefore, different private decryption exponent dy). With the knowledge of a single private decryption exponent d,, we can efficiently factor N , as explained below. Then the Euclidean Algorithm can then be used to recover all of the other decryption keys. Given a decryption exponent d and the corresponding public key ( e , N ) , we can determine the factors of N as follows [19]. First, we compute the number k = de - 1. Because of the way d and e are constructed, we know that k is a multiple of 4 ( N ) ,say, k = & ( N ) for some l. Since 4 ( N ) is even, so is k . Then k = 2tr, for some odd T and t 2 1. By a similar argument as that used above to show that the RSA algorithm works, we have
g k = g e d ( N )= 1 (mod N )
for every g E { 1 , 2 , . . . ,N - l} and, therefore, 91 is a square root of unity, modulo N , that is, (g / )* = 1 (mod N ) . The number 1 has four square roots, modulo N = pq. Two of these square roots are *1 and the other two (which can be found using the Chinese Remainder Theorenl-see Section A-2 in the Appendix), are fz, where z satisfies the conditions z = 1 (mod p ) and IC = -1 (mod 4 ) . Using either one of these last two square roots, the factorization of N is revealed by computing gcd(n:- 1,N ) . l A straightforward argument [19] shows that if g E { 1 , 2 , . . . , N - 1) is chosen at random, then with probability at least 1/2 one of the elements in the sequence