1. Prime Numbers: A Recent Discovery, Secure Communications, and Million Dollar Prizes George T. Gilbert TCU January 29, 2003

3. A prime number is a positive integer greater than 1 which is only divisible by 1 and itself. 2, 3, 5, 7, 11, 13, … are prime 1, 4=2  2, 6=2  3, 8=2  4, 9=3  3, 10=2  5, 12=3  4, … are not prime. 4, 6, 8, 9, 10, 12, … are called composite. The (approximate) definition of a prime number was first recorded by Philolaus (c. 480 - 390 B.C.), a member of the Pythagorean school.

4. Euclid (c. 325 - 265 B.C.): There are infinitely many primes. Proof. If not, list them out: p 1, p 2, …, p k Then p 1  p 2  …  p k +1 is 1 greater than a multiple of p 1, p 2, …, p k, so must be divisible by a prime not on the list.

5. The largest known prime is 2 13,466,917 -1, which has 4,053,946 digits, a result due to Michael Cameron (age 20, Canada), a “member” of the GIMPS (Great Internet Mersenne Prime Search) team Electronic Frontier Foundation Cooperative Computing Awards PrizeFirst Prime w/ $50,000 1,000,000 digits awarded April 6, 2000 $100,000 10,000,000 digits $150,000 100,000,000 digits $250,0001,000,000,000 digits

6. A First General Test for Primality Simply start checking for divisibility by 2, 3, 4, 5, 6, 7, … A number n is prime if it isn’t divisible by any number up to  n Note that once we find a factor, we could continue the procedure to get a factorization of n into a product of prime numbers. For example 200=2 3  5 2

7. COMPUTERS!!! At a billion divisions per second, this method would take 12 days to determine that a 30-digit number is prime and 32 trillion years to determine that a 60-digit number is prime. (The age of the universe seems to be 10-20 billion years.) Yet the state of the art is that one determine whether a 100-digit number is prime in seconds, several hundred digits in a few minutes, and Several 1000+ -digit numbers have been shown to be prime. Why the discrepancy? First, we chose a slow method, but it also appears that determining whether a number is prime is a much easier (=faster) question than factoring it. http://www.alpertron.com.ar/ECM.HTM

8. Fermat’s Little Theorem If p is prime, then for all k, k p -k is divisible by p. E.G. 3 4,294,967,297 -3 has remainder 497,143,886 upon division by 4,294,967,297, so 4,294,967,297 cannot be prime.

9. Public Key Cryptography and RSA In public key cryptography, one wishes to publish a key that allows anyone to encode a message but only the desired recipient to decode it (or vice versa). The RSA (Rivest-Shamir-Adelman, 1978) method is based on Fermat’s Little Theorem and depends on being able to find large primes quickly, whereas anyone given the product of two large primes “cannot” factor the number in a reasonable time.

10. Complexity of Algorithms We measure the complexity of a problem by how many steps it takes to solve the problem relative to the size of the input(s). For us, we will look at the worst case scenario, but in practice it also makes sense to look at an average case or most cases or something similar The size of a number n is number of digits (or bits in its binary representation), essentially log 10 n or log 2 n In this context, an algorithm is polynomial time if there exist constants C and m for which the algorithm finishes in at most C(log n) m steps

11. N10 3 10 6 10 12 10 20 10 50 10 200 10 1000 √N3210 3 10 610 10 25 10 100 10 500 log 5 N243 8  10 3 2  10 5 3  10 6 3  10 8 3  10 11 10 15 10 6  log 2 N 9  10 6 4  10 7 10 8 4  10 8 10 9 4  10 10 10 12

12. A Connection to The Riemann Hypothesis If p is prime and p divides k 2 -1=(k-1)(k+1), then p divides either k-1 or k+1. If p is not prime, then under a generalization of the Riemann hypothesis, we can find a k with p dividing k 2 -1 but dividing neither k+1 nor k-1 within a sufficiently small collection of k’s to give a polynomial time algorithm for primality testing (Miller 1975).

13. The Riemann hypothesis (G. F. Bernard Riemann, 1859), concerns the location of zeros of the zeta function, a meromorphic continuation of  n -s One fairly nontechnical equivalent condition is due to Jeff Lagarias (Amer. Math. Monthly 6/2002) Let H k = 1+1/2+1/3+…+1/k Then R.H. is equivalent to  d|k d ≤ H k +exp(H k ) ln(H k ) for all k with equality if and only if k=1 The Riemann hypothesis is one of the seven Millenium Prize Problems for which the Clay Mathematics Institute is offering $1,000,000 for a proof.*

14. Certificates of Primality This means a verification that may come out of the blue. There is such a quickly computed certificate for primality based on just a little more than Fermat’s Little Theorem: For each prime q that divides p-1, demonstrate a k for which k (p-1)/q -1 is not divisible by p and yet k p-1 -1 is divisible by p.

15. A problem is in the class … … NP (=non-deterministic polynomial time) if there exists a polynomial time method for verifying the answer. … P (=polynomial time) if there exists a polynomial time method for deriving the answer; P  NP This past summer it was proved that primality testing is in P. Factoring is in NP. If you could show factoring is not in P, you would have proved P≠NP and the Clay Math Institute would be writing your $1,000,000* check. Of course, “P=NP” and againP=NPagain

16. Theorem (M. Agrawal, N. Kayal, N. Saxena, 2002) There exists a polynomial time algorithm for testing primality. Preprint is only 8.5 pages Most of first 3 pages are history, summary, and preliminary facts from a standard graduate algebra course. Last 2 pages are a conjectural improvement in the exponent and the bibliography The rest is almost self-contained at the level of a graduate algebra course

17. The proof exploits the congruence (x-a) p  x p -a (mod p) It further reduces mod x r -1 for a cleverly chosen prime r of modest size The proof consists of elementary manipulations within finite fields and only calls on two outside results: (1/6) n / ln n ≤ #{primes ≤ n} ≤ 6 n / ln n Sieving result (Fouvry, 1985) For a positive density of primes p, the largest prime factor of p-1 is at least p 2/3