n - φ n + 1 = pq - (p 1)(q 1) + 1 = pq pq + p + q = p + q. = , then solving (by using the quadratic formula, TI-92, or Mathematica)
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1 Fall 006 Chis Chistensen MAT/CSC 483 Mathematical attack on RSA If we know and the public key (the modulus n and the encyption exponent e), then we can detemine d because d is the invese of e mod n We can use the Extended Euclidean algoithm (in Mathematica, ExtendedGCD[intege, intege]) to detemine e Then we can ead the message Now knowing is mathematically equivalent to knowing p and q the two pime factos of n Why? Well, cetainly is we know p and q we know = (p 1)(q 1) Convesely, if we know n and, then because n - φ n + 1 = pq - (p 1)(q 1) + 1 = pq pq + p + q = p + q So, we know n = pq and n - has p and q as its oots + 1 = p + q This suggests a quadatic equation which ( φ 1) X n n + X + n = X p X q Fo example, if we know that n = and = , then solving (by using the quadatic fomula, TI-9, o Mathematica) n = 7841*3463 So, knowing X X X X = and the public key (which allows us to beak any message encypted with that key) is equivalent to factoing n Thee is no known efficient algoithm to facto lage integes So, how might we facto n? We know that n is composite (in fact we know that it is the poduct of lage pimes) One of the factos of n must be less than o equal to division by all positive integes less than o equal to n The bute foce attack is to ty n This is not ecommended Thee is a factoing algoithm due to Femat ( ) that helps if the pimes ae nealy the same size Hee is how it woks
2 Femat factoing algoithm The algoithm is based upon the being able to facto the diffeence of squaes x y = x+ y x y If n x y, then n factos: n = x + y x y But, evey positive odd intege can be witten as the diffeence of two squaes In paticula fo the integes that we use of RSA moduli n = pq, = p + q p q n = pq = Let k be the smallest positive intege so that k > n, and conside k n If this is a squae, we can facto n: if k n = h, then n = ( k + h)( k h) If k n is not a squae, incease the tem on the left by one and conside ( 1) n factos If ( k + 1) n is not a squae, conside n k + n If this is a squae, k + Etc Eventually, we n+ 1 n 1 will find an h so that ( k + h) n factos That is so because n = n + 1 In this case, n factos as n= n 1 k k + h Hee is an example n = n ; so, k = 589 k n = = 58 So, = = = Femat s factoization algoithm woks well if the factos ae oughly the same size
3 Hee is anothe example n = ; so, k = is not a squae is not a squae is not a squae = 757 So, = = = The Pollad p 1 factoization algoithm Let s just conside the case of inteest factoing n = pq whee p and q ae lage pimes This algoithm woks well if eithe p 1 o q 1 is a poduct of elatively small pimes Let s assume that p 1 is the poduct of small pimes Fist, we guess an so that p 1 divides Of couse, in pactice we will not know p, but fo the moment assume that we do know p It might be convenient to take to be a factoial that is lage enough that p 1 divides Fo example, = 5 7 ; so, 15! Would wok because, 5, and 7 each divide ! = 3 7 ; so, 9! would wok But, = 3 163; so, we would need to use at least 163! This is what we mean by saying that we want to take = k! sufficiently lage so that p 1 divides But, emembe that because we won t know p we ae guessing that we have chosen lage enough If is not lage enough, the algoithm will fail to find a facto Assuming that p 1 divides, we can wite = (p 1) j Also choose a positive intege a so that 1 < a < p-1 Once again, we e guessing that ou a satisfies this inequality Say, we have sufficiently lage so that p 1 divides and a so that 1 < a < p 1 Then notice that ( p 1) j 1 p j a = a = a = 1 = 1mod p j
4 What this tells us is that p divides a 1, and because n = pq, p divides gcd a 1, n In fact, anothe a) ( a n) gcd 1, = p (Unless, by chance, a 1= 0modn In that case we choose So, that s the algoithm Afte choosing lage enough so that p 1 divides and a so that 1 < a < p, we calculate gcd a 1mod n, (We have good algoithms fo gcd and fo modula exponentiation) If we have chosen and a coectly, gcd a 1mod n, n = p Example: Let n = , a =, and = 13! PoweMod[, 13!, ] GCD[% - 1, ] 773 ( Which is one of the factos of n = = If we took too small hee s what happens Say, = 10! (Afte the fact we can see that = ; so 10! is too small) PoweMod[, 10!, ] GCD[% - 1, ] 1 n)
5 Factoing RSA s public key consists of the modulus n (which we know is the poduct of two lage pimes) and the encyption exponent e The pivate key is the decyption exponent d Recall that e and d ae inveses mod Knowing and n is equivalent to knowing the factos of n One attack on RSA is to ty to facto the modulus n If we could facto n, we could calculate φ n and (by using the extended Euclidean algoithm) detemine d Hee ae some factoing techniques: Tial division: Ty all the pimes that ae n It s not vey elegant, but in theoy it would wok The poblem is that, like othe bute foce techniques, it s not pactical Femat ( ) factoization: Not a bad technique if p and q ae elatively equidistant fom n (1974) Pollad p-1 algoithm: Not bad if p 1 o q 1 is the poduct of small pimes (1975) Pollad ρ -algoithm: The book discusses this algoithm (1981) Pomeance quadatic sieve algoithm QSA: Still fast fo up to aound 110 decimal digits (c 1993) Numbe field sieve NFS: The most efficient; based on wok of Pollad (1988) (c 1987) Elliptic cuve method ECM: H Lensta (1994) Scho s algoithm: Needs a quantum compute The existing 7 cubit quantum compute has factoed 15 The RSA factoing challenge numbes
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