Sifting the Primes. Gihan Marasingha University of Oxford

Transcription

1 Sifting the Primes Gihan Marasingha University of Oxford 18 March 2005

2 Irreducible forms: q 1 (x, y) := a 1 x 2 + 2b 1 xy + c 1 y 2, q 2 (x, y) := a 2 x 2 + 2b 2 xy + c 2 y 2, a i, b i, c i Z. Variety V defined by: V : q 1 (x, y) = u 2 + v 2 q 2 (x, y) = s 2 + t 2 1

3 Comp osite numb ers n. If n is comp osite, then it has a prime factor p with p n : Thus, having struck out multiples of primes p, we've extinguished all the comp osite numb ers. The Sieve of Eratosthenes

4 The Sieve of Eratosthenes Prime numbers n. If n is composite, then it has a prime factor p with p n. Thus, having struck out multiples of primes p, we ve extinguished all the composite numbers less that. 2-a

5 The Sieve of Eratosthenes Legendre π() := the number of primes p. S(, r) := #{n : 2, 3, 5,..., p r n}, where p r < is the r-th prime. Then we have the relationship: π() p r + S(, r). Reason: suppose p counted by π(), so that p. Either p p r or p > p r. In the first case, p {1,..., p r }, so p is counted by p r, otherwise p is counted by S(, r). Define k := #{n : k n}, then S(, r) = pr pi p j pi p j p k +... ± p1...p r 3

6 ow k counts k, 2k, 3k,... qk where qk < (q + 1)k, so k k < k + 1. S(, r) = p r p i p j p i p j p k +... ± p 1... p r + error, where error 2 r 2 p r. We ll write error= O(2 p r), where f(n) = O(g(n)) means that there exists a constant C such that f(n) Cg(n). In our case C = 1. So S = i r p i + O(2 pr ). ) (1 1pi = r i=1 i j r + O(2 pr ) 1 p i p j... ± 1 p 1... p r otation: f(n) g(n) means that f(n)/g(n) 1 as n. 4

7 Fact: there is a constant C such that p<z ( 1 1 p ) C 1 log z. Choose r such that p r < log p r+1, then S(, r) = p<log ( 1 1 p ) C log log + log 2 = O( log log ) = O( log log ). Thus: π() S(, r) + p r = O( log log = O( log log ). + O(2 log ) + log ) Theorem (Hadamard, de la Vallée Poussin, 1896). π() log. 5

8 Theorem (Hadamard, de la Vallée Poussin, 1896). π() log. Conjecture (Goldbach, 1750). Let be an even number greater that 2, then = p + q, for some primes p and q. How many representations? That is what is Heuristically, it s p p #{p : p is prime}? prob. that p is prime p 1 log( p) 1 log 1 log π() 1 log log = (log ) 2. 6

9 Definition: We say n is a k-almost prime and write that n is P k if n has at most k prime factors. Theorem (Chen Jing-Run, 1974). For all sufficiently large even, one has that = p + P 2, for p a prime. More precisely, there exists a (computable) constant C such that for all sufficiently large even, {p : p, p = P 2 } > C (log ) 2. Idea: Get a good lower bound for representations = p+p 3 then take away the representations = p + p 1 p 2 p 3 by deducing an upper bound. What s left are the representations = p + P 2. 7

10 Essentially, Chen is interested in calculating a lower bound for the number of 2-almost primes in the set A := { p : p }, much as in our heuristic development of the Goldbach conjecture. Chen s primary innovation in the solution of this problem was the reversal of rôles, with which he relates A to B, where B := { p 1 p 2 p 3 : p 1 p 2 p 3 <, 1/10 p 1 < 1/3 p 2 < p 3 }, so that B refers to the number of representations of p as the sum of a prime and a product of exactly three primes. It is an upper bound for the number of primes in B which is taken away from the number of representations = p + P 3 to provide our lower bound for the number of almost prime in A. 8

11 Conjecture (Twin Primes). There exist infinitely many primes p such that p+2 is prime. Theorem (Chen Jing-Run, 1974). Let h be an even natural number. Then there exist infinitely many primes p such that p + h = P 2. Theorem (Brun, 1912). The sum p p+2 prime 1 p + 1 p + 2 is convergent, its value being referred to as Brun s constant. 1995: Thomas icely computed prime twins up to 10 14, and found a bug in the Intel Pentium! 9

12 Conjecture (Euler, 1752). There exist infinitely many primes p of the form p = x Theorem (Dirichlet, 1837). If a, b are coprime integers, then there exist infinitely many primes p of the form p = ax + b. Hypothesis H (Schinzel, Sierpinski, 1958). Let F 1 (x),..., F n (x) be distinct irreducible polynomials with integer coefficients, then under a certain condition on the product, there exist infinitely many x such that each F i (x) is prime. 10

13 Eratosthenes Legendre: primes in A := {n : n }. Introduced k := A k with A k := {n : k n}. Approximated k by /k and found R k = k /k is bounded by R k 1. Theorem (Halberstam & Richert, 1972). Let A be a set of integers with A X. Then, under certain conditions, one can find constants r, κ, δ > 0 and C 1 such that X {P r : P r A} δ (log X) κ ( 1 ) C. log X The crucial condition in the determination of the least number of almost primes r is a good bound for the error term R k. We look for a condition something like: d<x α R k C X (log X) κ. If we can find an estimate with a large value of α, then we may correspondingly use a small value of r. 11

14 Theorem (H&R, 1972). Let Q 1, Q 2 be irreducible quadratic polynomials over the integers such that Q 1 Q 2 has no fixed prime divisor, then there exist infinitely many integers n such that Q 1 (n)q 2 (n) = P 9. Theorem (Iwaniec, 1972). Let F (x, y) be a quadratic polynomial. Then, under a certain simple condition on the coefficients, the number of primes p represented by F (x, y) is of order /(log ) 3/2. 12

15 Theorem (M, 2005). Let q 1, q 2 be irreducible binary quadratic forms over the integers, then subject to certain conditions on the forms, then there exist infinitely many pairs of integers (n, m) such that q 1 (n, m)q 2 (n, m) = P 6. Old problem: investigate the variety V : q 1 (x, y) = u 2 + v 2 q 2 (x, y) = s 2 + t 2 Count (X), the points (x, y, u, v, s, t) Z 6 with x, y < X, and derive asymptotic formula as X. In calculating the asymptotic formula, one needs to evaluate quantities of the type k<x R k, as in Halberstam and Richert s theorem. 13

16 More Questions: Extend results of Marasingha? Pairs of irreducible cubic forms? Triples of quadratic forms? There are sieve methods for calculating π() in time O( 2/3+ɛ ), which don t require the computation of all the primes, but it seems that to compute π 2 () := the number of twin primes, we need to compute all the twin primes, taking time O(). Can we improve on this? 14