Simultaneous Prime Values of Two Binary Forms

Transcription

1 Peter Cho-Ho Lam Department of Mathematics Simon Fraser University

2 Twin Primes Twin Prime Conjecture There exists infinitely many x such that both x and x + 2 are prime. Question 1 Are there infinitely many x, y such that both x and x + 2y are primes?

3 Twin Primes Twin Prime Conjecture There exists infinitely many x such that both x and x + 2 are prime. Question 1 Are there infinitely many x, y such that both x and x + 2y are primes?

4 Main Problem Question 2 Let F, G [x, y] be two irreducible binary forms. Are there infinitely many x, y such that both F(x, y) and G(x, y) are primes? Fouvry and Iwaniec (1997) There are infinitely many primes of the form x 2 + y 2 such that y is also a prime. Main Result Let F, G [X, Y] be two irreducible binary forms such that deg F = 2 and deg G = 1. If F is positive definite, then there are infinitely many x, y such that F(x, y) and G(x, y) are both primes.

5 Main Problem Question 2 Let F, G [x, y] be two irreducible binary forms. Are there infinitely many x, y such that both F(x, y) and G(x, y) are primes? Fouvry and Iwaniec (1997) There are infinitely many primes of the form x 2 + y 2 such that y is also a prime. Main Result Let F, G [X, Y] be two irreducible binary forms such that deg F = 2 and deg G = 1. If F is positive definite, then there are infinitely many x, y such that F(x, y) and G(x, y) are both primes.

6 Main Problem Question 2 Let F, G [x, y] be two irreducible binary forms. Are there infinitely many x, y such that both F(x, y) and G(x, y) are primes? Fouvry and Iwaniec (1997) There are infinitely many primes of the form x 2 + y 2 such that y is also a prime. Main Result Let F, G [X, Y] be two irreducible binary forms such that deg F = 2 and deg G = 1. If F is positive definite, then there are infinitely many x, y such that F(x, y) and G(x, y) are both primes.

7 Sieve Method - Asymptotic Sieve We wish to obtain an asymptotic formula for the sum a n Λ(n) where Λ(n) is the von Mangoldt function, log p if n = p k, k, Λ(n) = 0 otherwise. For example, we can take a n = 1 or a n = Λ(n + 2).

8 Sieve Method - Asymptotic Sieve We wish to obtain an asymptotic formula for the sum a n Λ(n) where Λ(n) is the von Mangoldt function, log p if n = p k, k, Λ(n) = 0 otherwise. For example, we can take a n = 1 or a n = Λ(n + 2).

9 Asymptotic Sieve In general, since µ(d) log d = Λ(n) d n where µ is the Möbius function, µ(n) = ( 1) r 0 otherwise, if n = p 1 p 2...p r, p i are all distinct primes we have a n Λ(n) = a n µ(d) log d = µ(d) log d d n d x n 0 (mod d) a n.

10 Asymptotic Sieve In general, since µ(d) log d = Λ(n) d n where µ is the Möbius function, µ(n) = ( 1) r 0 otherwise, if n = p 1 p 2...p r, p i are all distinct primes we have a n Λ(n) = a n µ(d) log d = µ(d) log d d n d x n 0 (mod d) a n.

11 Asymptotic Sieve Suppose for all d we have the following approximation: a n = g(d) a n + r d (x) n 0 (mod d) for some multiplicative function g(d). Then the sum we need to estimate turns into µ(d) log d d x n 0 (mod d) a n = µ(d) log d g(d) d x µ(d)g(d) log d a n. d x a n + r d (x)

12 Asymptotic Sieve Suppose for all d we have the following approximation: a n = g(d) a n + r d (x) n 0 (mod d) for some multiplicative function g(d). Then the sum we need to estimate turns into µ(d) log d d x n 0 (mod d) a n = µ(d) log d g(d) d x µ(d)g(d) log d a n. d x a n + r d (x)

13 ASYMPTOTIC Sieve For nice functions g (say g(d) = 1/d), we have µ(d)g(d) log d = (1 g(p)) = H. p d p Therefore if the remainder terms r d (x) are small on average, then we expect a n Λ(n) H for some constant H. When a n = 1, we have g(d) = 1/d, H = 1, and we get back the Prime Number Theorem. a n

14 ASYMPTOTIC Sieve For nice functions g (say g(d) = 1/d), we have µ(d)g(d) log d = (1 g(p)) = H. p d p Therefore if the remainder terms r d (x) are small on average, then we expect a n Λ(n) H for some constant H. When a n = 1, we have g(d) = 1/d, H = 1, and we get back the Prime Number Theorem. a n

15 Type I: we wish to show that r d (x) o a n d D for D as large as possible (best possible: D = x 1 ε ).

16 Type II: for large values of d, µ(d) log d D<d x n 0 (mod d) a n = m D<m x µ(m)(log m)a mn. The logarithm factor can be removed and it suffices to estimate µ(m)a mn. n N m M

17 Settings Let F(x, y) = αx 2 + βxy + γy 2 and a n = l,m F(l,m)=n (l,m)=1 Λ(m). Two goals: estimate a n g(d) a n d D n 0 (mod d) and µ(m)a mn. n N m M

18 A d (f) = = F(l,m) 0 (mod d) (l,m)=1 ν (mod d) a F(ν,1) 0 (mod d) Λ(m)f(F(l, m)) µ(a) Λ(am) m l l νm (mod d) By Poisson summation formula, the inner sum becomes f(f(al, am)) = 1 hνm h e F a,m d d d h where l νm (mod d) F a,m (z) = f(f(at, am))e( zt) dt. f(f(al, am)).

19 Large sieve: if d 1, d 2 D, F(ν 1, 1) 0 (mod d 1 ), F(ν 2, 1) 0 (mod d 2 ), then ν 1 ν 2 d 1 1 D. Balog, Blomer, Dartyge and Tenenbaum (2011) Let F(x, y) = αx 2 + βxy + γy 2 [x, y] be an arbitrary quadratic form whose discriminant = β 2 4αγ is not a perfect square. For any sequence α n of complex numbers, positive real numbers D, N, we have D d 2D F(ν,1) 0 (mod d) n N d 2 νn 2 α n e F (D + N) α n 2. d n

20 Large sieve: if d 1, d 2 D, F(ν 1, 1) 0 (mod d 1 ), F(ν 2, 1) 0 (mod d 2 ), then ν 1 ν 2 d 1 1 D. Balog, Blomer, Dartyge and Tenenbaum (2011) Let F(x, y) = αx 2 + βxy + γy 2 [x, y] be an arbitrary quadratic form whose discriminant = β 2 4αγ is not a perfect square. For any sequence α n of complex numbers, positive real numbers D, N, we have D d 2D F(ν,1) 0 (mod d) n N d 2 νn 2 α n e F (D + N) α n 2. d n

21 We need to estimate µ(m)a mn. n N m M What do we know about m, n if for some x, y, (x, y) = 1? mn = F(x, y) = αx 2 + βxy + γy 2

22 We need to estimate µ(m)a mn. n N m M What do we know about m, n if for some x, y, (x, y) = 1? mn = F(x, y) = αx 2 + βxy + γy 2

23 Simple example: F(x, y) = x 2 + y 2. If m = a 2 + b 2 and n = u 2 + v 2, then mn = x 2 + y 2 with x = au + bv, y = av bu since (a 2 + b 2 )(u 2 + v 2 ) = (au + bv) 2 + (av bu) 2. Conversely, if mn = x 2 + y 2 with (x, y) = 1, then we can write m = a 2 + b 2 and n = u 2 + v 2 such that x = au + bv, y = av bu. It is NOT true in general.

24 Simple example: F(x, y) = x 2 + y 2. If m = a 2 + b 2 and n = u 2 + v 2, then mn = x 2 + y 2 with x = au + bv, y = av bu since (a 2 + b 2 )(u 2 + v 2 ) = (au + bv) 2 + (av bu) 2. Conversely, if mn = x 2 + y 2 with (x, y) = 1, then we can write m = a 2 + b 2 and n = u 2 + v 2 such that x = au + bv, y = av bu. It is NOT true in general.

25 Simple example: F(x, y) = x 2 + y 2. If m = a 2 + b 2 and n = u 2 + v 2, then mn = x 2 + y 2 with x = au + bv, y = av bu since (a 2 + b 2 )(u 2 + v 2 ) = (au + bv) 2 + (av bu) 2. Conversely, if mn = x 2 + y 2 with (x, y) = 1, then we can write m = a 2 + b 2 and n = u 2 + v 2 such that x = au + bv, y = av bu. It is NOT true in general.

26 Simple example: F(x, y) = x 2 + y 2. If m = a 2 + b 2 and n = u 2 + v 2, then mn = x 2 + y 2 with x = au + bv, y = av bu since (a 2 + b 2 )(u 2 + v 2 ) = (au + bv) 2 + (av bu) 2. Conversely, if mn = x 2 + y 2 with (x, y) = 1, then we can write m = a 2 + b 2 and n = u 2 + v 2 such that x = au + bv, y = av bu. It is NOT true in general.

27 Trickier example: F(x, y) = x 2 + 5y 2 and (a 2 + 5b 2 )(u 2 + 5v 2 ) = (au + 5bv) 2 + 5(av bu) 2. For example, 2 3 = (1) 2 + 5(1) 2, but both 2 = x 2 + 5y 2 and 3 = x 2 + 5y 2 are not even solvable in integers. But 2 and 3 can be represented by 2x 2 + 2xy + 3y 2, and we also have the identity (2a 2 +2ab+3b 2 )(2u 2 +2uv+3v 2 ) = (2au+av+bu+3bv) 2 +5(au bv) 2.

28 Trickier example: F(x, y) = x 2 + 5y 2 and (a 2 + 5b 2 )(u 2 + 5v 2 ) = (au + 5bv) 2 + 5(av bu) 2. For example, 2 3 = (1) 2 + 5(1) 2, but both 2 = x 2 + 5y 2 and 3 = x 2 + 5y 2 are not even solvable in integers. But 2 and 3 can be represented by 2x 2 + 2xy + 3y 2, and we also have the identity (2a 2 +2ab+3b 2 )(2u 2 +2uv+3v 2 ) = (2au+av+bu+3bv) 2 +5(au bv) 2.

29 Trickier example: F(x, y) = x 2 + 5y 2 and (a 2 + 5b 2 )(u 2 + 5v 2 ) = (au + 5bv) 2 + 5(av bu) 2. For example, 2 3 = (1) 2 + 5(1) 2, but both 2 = x 2 + 5y 2 and 3 = x 2 + 5y 2 are not even solvable in integers. But 2 and 3 can be represented by 2x 2 + 2xy + 3y 2, and we also have the identity (2a 2 +2ab+3b 2 )(2u 2 +2uv+3v 2 ) = (2au+av+bu+3bv) 2 +5(au bv) 2.

30 In general, if mn = F(x, y) and (x, y) = 1, then we wish to show that m can be represented by another binary quadratic form of the same discriminant, say f. And then n can be represented by g, where f g = F, g = F f 1. We would also need a precise formula for g and an identity for the convolution.

31 In general, if mn = F(x, y) and (x, y) = 1, then we wish to show that m can be represented by another binary quadratic form of the same discriminant, say f. And then n can be represented by g, where f g = F, g = F f 1. We would also need a precise formula for g and an identity for the convolution.

32 Factorization Lemma If mn = αx 2 + βxy + γy 2 for some integers X, Y with (X, Y) = 1, then there exists a binary quadratic form f(x, y) = ax 2 + bxy + cy 2 and integers u, v, w, z such that (u, v) = (w, z) = 1 and au 2 + buv + cv 2 = m, aαw 2 + Bwz + B2 + 4aα z2 = n, au + b + β B β (b + β)b + bβ v w + u + v z = X, 2 2α 4aα αvw + u B b 2a v z = Y; and the choice of f, u, v, w, z are "unique".

33 Related Problems 1 quadratic+linear 2 cubic+linear 3 quadratic + quadratic

34 THE END!