This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PMATH 440/640 ANALYTIC NUMBER THEORY 1. Lecture: Monday, September 11, 2000 NO TEXT References: Introduction to Analytic Number Theory. Tom Apostol, SpringerVerlag, 1976 An Introduction to the Theory of Numbers. G. H. Hardy and E. M. Wright, Oxford Uni versity Press (5th ed, 1979) Marks: Final Exam 65% Midterm 25% Assignments 10% Is there a formula for n th prime p n ? p 1 = 2 , p 2 = 3 , . . . Yes, but such a formula is complicated. For example, is there a polynomial f Z [ x ] for which f ( n ) = p n ? f ( x ) = a n x n + + a 1 x + a f ( a ) = a n a n + + a 1 a + a so a  f . Suppose q is prime and f ( n ) = q . Then q  f ( n + kq ) for each k Z + . So, in particular, we see that if f ( m ) is prime for each positive integer m , then f is a constant. In particular, f ( x ) = q for some prime q . The polynomial n 2 + n + 41 is prime for n = 0 , 1 , . . ., 39. Further, ( n 40) 2 + ( n 40) + 41 is prime for 0 n 79. These examples are connected with the fact that the ring of algebraic integers of the field Q ( 163) is a Unique Factorization Domain. (Note that 163 is the largest squarefree integer D such that Q ( D ) is a U.F.D.) By using ideas of Matijasevic which were used to prove Hilberts 10th problem, one can find a polynomial f Z [ a, b, . . ., z ] such that the set of positive values assumed by f as the variables run over the nonnegative integers is the set of prime numbers. Is n 2 + 1 a prime for infinitely many n ? Almost surely yes. The best result in this direction is that n 2 +1 is a P 2 for infinitely many integers n . A P 2 is an integer which is the product of at most 2 primes. 1 2 PMATH 440/640 ANALYTIC NUMBER THEORY There is no polynomial of degree > 1 which is known to be prime infinitely often. On the other hand, we can deal with degree 1. Let k and be coprime positive integers. Then kn + is prime for infinitely many positive integers n . This is Dirichlets Theorem . Theorem 1 (Euclid) . There are infinitely many prime numbers. Proof. Assume that there are finitely many primes p 1 , . . ., p n say, and consider m = p 1 p n + 1 Then m can be written as a product of primes, and so p k  m for some k with 1 k n . Then p k  m p 1 p n , so p k  1, which is a contradiction. square Definition. For any real number x , let ( x ) denote the number of primes x . We can estimate ( x ) from below using Euclids proof. In particular, well show that p k , the k th prime, satisfies p k 2 2 k for k = 1 , . . ., n We prove this by induction. The result holds for k = 1, since 2 = p 1 2 2 = 4. Assume the result holds for 1 j k ....
View
Full
Document
 Fall '00
 NoneListed

Click to edit the document details