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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, Springer-Verlag, 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 non-negative 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 ....
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