Primality Testing

1. Primality Testing Patrick Lee 12 July 2003 (updated on 13 July 2003)

2. Finding a Prime Number • Finding a prime number is critical for public-key cryptosystems, such as RSA and Diffie-Hellman. • Naïve approach: • Randomly pick a number n. Try if n is divided by 2, 3, 5, 7, …., p, where p is the largest prime number less than or equal to the square root of n. • Computationally expensive. • You need to pre-obtain all small prime numbers.

3. Introduction to Number Theory • Number theory: modular arithmetic on a finite set of integers • Most of the randomized algorithms starts by choosing a random number from some domain and then works deterministically from there on. We hope that with high probability the chosen number has some desirable properties. • Goal: Given a number n, the desired complexity is O(logn), i.e., polynomial in the length of n.

4. Computing GCD • gcd(a, b): greatest common divisor of (a,b) • a and b are co-prime iff gcd(a,b) = 1 • Euclid’s algorithm: • Finding gcd(a,b) • for a>b, gcd(a,b) = gcd(b, a mod b) • Extended Euclid’s: • Finding gcd d and numbers x and y such that d=ax+by

5. Groups • Additive Group: • Zn= {0, 1, …, n-1} forms a group under addition modulo n. • Multiplicative Group: • Zn* = {x | 1 <= x < n and gcd(x,n) = 1} forms a group under multiplication modulo n. • For prime p, Zp* includes all elements [1,p-1]. • E.g., Z6* = {1, 5} • E.g., Z7* = {1, 2, 3, 4, 5, 6}

6. Chinese Remainder Theorem (CRT) • Given n1, n2,…, nk are pairwise co-prime. There exists a unique r, r in [0, n = n1n2…nk), satisfying r = ri mod ni for any sequence {r1,..,rk}, where ri in [0, ni). • E.g., r = 2 (mod 3) r = 3 (mod 5) r = 2 (mod 7) We have r = 23, unique in [0,105).

7. Euler phi Function: phi(.) • phi(n) = |Zn*| • e.g., phi(p) = p–1 for prime p • Theorem: if n= p1e1p2e2…pkek, phi(n) = (p1-1)p1e1 - 1...(pk-1)pkek – 1 • e.g., if n = pq, phi(n) = (p-1)(q-1) • If we know phi(n), we can factorize n. • Euler’s Theorem: for all n and x in Zn* xphi(n) = 1 (mod n) • For any prime p, xp-1 = 1 (mod p) for all x in [1, p-1]. (Fermat’s Little Theorem). • If xn-1 <> 1, n is not prime (e.g., 45 mod 6 = 4).

8. Order and Generator • ord(x): smallest t such that xt = 1 mod n • E.g., in Z11*, ord(3) = 5, ord(2) = 10 • Generator: an element whose order = group size. • E.g., 3 is the generator of Z7* • Subgroup: generated from an element of order t < phi(n) • {1,3,32=9,33=5,34=4} = {1,3,4,5,9} is a subgroup of Z11* • A group is cyclic if it has a generator. • For any prime p, the group Zp* is cyclic, i.e, every Zp* has a generator, say g. • Zp* = {1, g, g2, g3, …, gp-2}

9. Group Size • Subgroup size divides group size (for all n) • Group size = phi(n) • We use an element of order t < phi(n) as the generator of the subgroup, (say 2 in Z7*). • The subgroup spans t elements. • For x in subgroup, we observe t has to divide phi(n) so that xtk = xphi(n) = 1, for some integer k. You can prove it by contradiction by assuming t does not divide phi(n). • E.g., H = {1, 3, 4, 5, 9} is a subgroup of Z11*, |H| dividies |Z11*|. • This proposition applies to all n (prime / composite).

10. Quadratic Residue • y is a quadratic residue (mod n) if there exists x in Zn* such that x2 = y (mod n) • i.e., y has a square root in Zn* • Claim:For any prime p, every quadratic residue has exactly two square roots x, -x mod p. • Proof: if x2 = u2 (mod p), then (x-u)(x+u) = 0 (mod p), so either p divides x-u (i.e., x=u), or p divides x+u (i.e., x=-u). • It implies if x2 = 1 (mod p), x = 1 or -1.

11. Quadratic Residue (cont’d) • Theorem: For any prime p, and g is generator, gk is a quadratic residue iff k is even. • Given Zp* = {1, g, g2, g3, …, gp-2} • Even powers of g are quadratic residues • Odd powers of g are not quadratic residues • Legendre symbol: • [a/p] = 1 if a is a quadratic residue mod p, and -1 if a is not a quadratic residue mod p.

12. Quadratic Residue (cont’d) • Theorem: For prime p and a in Zp*, [a/p] = a(p-1)/2 (mod p). • Zp* is cyclic, a = gk for some k. • If k is even, let k = 2m, a(p-1)/2 = g(p-1)m = 1. • If k is odd, let k = 2m+1, a(p-1)/2 = g(p-1)/2 = -1. Reasons: • This is a square root of 1. • g(p-1)/2 <> 1 since ord(g) <> (p-1)/2. • But 1 has two square roots. Thus, the only solution is -1. • If n is prime, a(n-1)/2= 1 or -1. If we find a(n-1)/2 is not 1 and -1, n is composite.

13. Ideas of Primality Testing • Idea 1: • If xn-1 mod n <> 1, n is definitely composite. • If xn-1 mod n = 1, n is probably prime. • Idea 2: • If x(n-1)/2 mod n <> {1,-1}, n is definitely composite. • If x(n-1)/2 mod n = {1,-1}, n is probably prime.

14. Simple Primality Testing Alg. • Repeat k times: • Pick a in {2,...,n-1} at random. • If gcd(a,n) != 1, then output COMPOSITE. [this is actually unnecessary but conceptually helps] • If a(n-1)/2 is not congruent to +1 or -1 (mod n), then output COMPOSITE. Now, if we ever got a "-1" above output "PROBABLY PRIME" else output "PROBABLY COMPOSITE".

15. Error of the Simple Alg. • The alg is BPP with error probability 1/2k. • If n is prime, half of them makes a(n-1)/2 = 1. Prob. error in each iteration is ½. • If n is composite, error occurs if n is claimed to be “PROBABLY PRIME”. We use the key lemma. • Key Lemma: Let n be an odd composite, not a prime power, and let t=(n-1)/2. If there exists a in Zn* such that at = -1 (mod n), then at most half of the x's in Zn* have xt = {-1,+1} (mod n).

16. Error of the Simple Alg. (cont’d) • Let S = {x in Zn* | xt = 1 or -1} (let t = (n-1)/2). • We’d like to show S is a proper subgroup of Zn*. • S is a subgroup of Zn* since it's closed under multiplication (xt)(yt) = (xy)t. • Find b in Zn* but not in S. • Let n = qr, where q and r are co-prime. • Using the CRT notation, let b = (a,1), denoting b=a (mod q), b=1 (mod r). CRT assures the existence of b. • Thus, bt = (at, 1t) = (-1, 1), implying b <> 1 and -1, since 1 = (1, 1) and -1 = (-1,-1). • S is a proper subgroup. Since the subgroup size divides the group size, |S| <= ½ |Zn*|.

17. Case of Prime-Power Composites • Key Lemma doesn’t apply if n is a prime-power. However, it doesn’t matter since it cannot pass the test of step (3), i.e., we are sure that a(n-1)/2 <> 1,-1 mod n for all a. • Proof (assume all operations are mod n): • Write n = pe, where p is prime. • Consider an-1, which is equal to ape-1. • Note that phi(n) = pe-1(p-1) = pe-pe-1, according to the theorem in slide 7. • ape-1 = aphi(n)+pe-1-1 = ape-1-1 (by Euler’s Theorem) • Recursively, we get ape-1 = a-1. • Since a<>1, a-1 <> 1. We have an-1 <> 1, and its square root is not 1 and -1. • Thus, if n is prime-power, it does not pass the test case in step (3). We can safely ignore the case of prime-powers in the Key Lemma.

18. Miller-Rabin Algorithm • pick a in {2,...,n-1} at random. • If an-1 != 1 (mod n), then output COMPOSITE • Let n-1 = 2r * B, where B is odd. • Compute aB, a2B, ..., an-1 (mod n). • If we found a non {-1,+1} root of 1 in the above list, then • output COMPOSITE. • else output POSSIBLY PRIME.

19. Error of MR Algorithm • It is RP. • For prime n, the algorithm always returns prime. • For non-Carmichael composite n, the algorithm returns prime with probability at most ½ in each iteration (i.e., step 2 detects compositeness with probability at least ½). • Carmichael number: a composite n such that for all a in Zn*, an-1 = 1 mod n. (e.g., 561, 1729)

20. Error of MR Algorithm (Proof) • Let Fn = {x in Zn* | xn-1 = 1 mod n}, the set of elements that do not violate Fermat’s theorem. • Lemma: Let n be a composite non-Carmichael number. Then |Fn| <= ½ |Zn*|. • Clearly, Fn <> Zn* . • There exists a such that an-1 <> 1 mod n. • Fn forms a group. • It is closed under multiplication (trivial proof!) • Fn is a proper subgroup of Zn*. |Fn| divides |Zn*|, and |Fn| is strictly less than |Zn*|.

21. Detecting Carmichael Numbers • Computing aB, a2B, ..., a2rB (mod n), where B =(n-1)/2r, detects Carmichael numbers. • Idea: a(n-1)/2 = {1,-1}, how about a(n-1)/4? If a(n-1)/4 = {1,-1}, how about a(n-1)/8? • Prove by contradiction. • Assume n is Carmichael, for all a, aB = 1 mod n. • Property: Carmichael number is the product of distinct prime. Thus, let n = p1p2..pk. • Let g’ is a generator of Zp1*. • Let a = (g’, 1), i.e., a = g’ (mod p1), a = 1 (mod p2..pr), by CRT • By assumption, aB = 1 (mod n). It implies g’B = 1 (mod p1) (why?). • Since g’ is the generator, B = p-1, which contradicts B is odd. • Thus, for some a, aB <> 1. The probability is > ½.

22. How to Find a Prime Number? • Algorithm: • Randomly pick a number from [1,n-1]. • Plug it into the primality testing algorithm. • If fails, repeat the test with another number. • Are prime numbers rare? No. • Prime number theorem: • No. of prime numbers less than n ~ n/ln(n).

23. References • R. Motwani and P. Raghavan, “Randomized Algorithms”, Ch. 14. • CMU, “Randomized algorithms”, http://www-2.cs.cmu.edu/afs/cs/usr/avrim/www/Randalgs98/home.html • CLRS, “Introduction to Algorithms”, 2nd edition. Ch. 31.