On the Representation of Almost Primes by Pairs of Quadratic Forms

 Alban Gordon
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1 On the Reresentation of Almost Primes by Pairs of Quadratic Forms Gihan Marasingha May 27, 2008 Mathematics Subject Classification: Primary: N36, Secondary: N05 Introduction An integer is said to be an almost rime of order r, and is denoted P r, if it is the roduct of at most r (not necessarily distinct) rime factors. Schinzel s celebrated hyothesis H may be reformulated in the language of almost rimes as follows: Hyothesis H. Let F,..., F n be irreducible olynomials over the integers such that the roduct F := F...F n has no fixed rime divisor (that is, there does not exist a rime such that F (x) 0 (mod ) for all x N). Then there exist infinitely many x N such that F (x) = P n. The only verified case is that of one linear olynomial. This is Dirichlet s Theorem on arithmetic rogressions. As far as quadratic olynomials are concerned, one of the best results is due to Iwaniec [9], who modified a weighted linear sieve of Richert to demonstrate that x 2 + = P 2 for infinitely many x. It has been known since the time of Dirichlet which binary quadratic forms reresent rimes, see the books by Buell [] and Cox [2], for examle. However the situation for airs of forms aears to be comletely oen. In this aer, we investigate an aroximation to Schinzel s hyothesis for the case n = 2. The result we achieve involves binary quadratic forms rather This aer was written with the suort of an EPSRC studentshi, award number
2 than olynomials, and we shall show that the roduct of the forms is P 5 for infinitely many values of the variables, as oosed to the P 2 result redicted by Schinzel s hyothesis. Our resent theorem is an imrovement on Diamond and Halberstam s P 7 result for quadratic olynomials, a secial case of a result in [5]. To be recise, our main theorem is the following: Theorem.. Let q i (x, y) := a i x 2 + 2b i xy + c i y 2, for i =, 2, be irreducible quadratic forms over the integers such that a i (mod 4). Let δ i be the discriminant of the form q i. Let D := 6 Res(q, q 2 )a a 2 c c 2 δ δ 2, where Res(q, q 2 ) is the resultant of the forms q and q 2. If D 0 and if there exists z Z 2 such that (q i (z); D) = for i =, 2, then there exist infinitely many airs (x, y) Z 2 such that q (x, y)q 2 (x, y) = P 5. Moreover, if R (0) is a convex subset of R 2 with iecewise continuously differentiable boundary, then there exists a ositive absolute constant β < such that that for all sufficiently large X, #{(x, y) XR (0) : q (x, y)q 2 (x, y) = P 5 } X ( 2 ω() ), <X β where the imlied constant deends at most on the forms q and q 2, and on the region R (0) ; and where ω() = 2 + χ () + χ 2 () ( + χ () + χ 2 ())/, the characters χ and χ 2 defined by χ i () := ( δ i ). Our rincial external tool will be a multi dimensional sieve of Diamond and Halberstam [5], a secial case of which is resented as Theorem 3. in the resent work. The most exacting asect of our almostrimes roblem is the derivation of an uer bound for the error term R d which aears in Diamond and Halberstam s sieve. We devote Section 2 to the necessary groundwork. In doing so, we develo a level of distribution formula which is strongly related to the sum of the R d. Similar formulæ have been alied by HeathBrown [6], Daniel [3], and others in the investigation of asymtotic formulæ for the number of oints of bounded height on given varieties. We shall use the following standard notation: d(n) is the number of ositive divisors of n; φ(n) is the number of nonnegative integers less than and rime to n; µ(n) is the Möbius function; ν(n) is the number of distinct rime factors of n; and ( ) is the Legendre symbol. We will use the symbol C to denote a ositive numerical constant, though its value may vary in the course of a roof. 2
3 2 The Level of Distribution In our alication of Diamond and Halberstam s sieve, we shall encounter a sum over error terms R d, which we will be able to relate to quantities of the form #(Λ d R Ψ) ρ (d, d 2 ) (d d 2 D) 2 vol(r), where Λ d is a latticelike object, and (d d 2 ) 2 /ρ (d, d 2 ) lays a rôle similar to the determinant of the lattice. Naturally, we wish to derive a good uer bound for the error term involved. In the language of sieve theory, such a result is often referred to as a level of distribution formula. The sets Λ d and Λ d which we are concerned with are as follows: () Λ d := {x Z 2 : d i q i (x) (i =, 2)}, Λ d := {x Λ d : (x; d d 2 ) = }, where we use (a; b) to denote the highest common factor of a and b. A technicality arises due to roblems for those x such that (q i (x); D) >, with D as in Theorem.. This leads us to consider the set (2) Ψ := {x Z 2 : x z (mod D)}, and hence the task of estimating #(Λ d R Ψ). Define the multilicative functions ρ and ρ by: (3) ρ(d) := #{x [0, d d 2 ) 2 : d i q i (x) (i =, 2)}, ρ (d) := #{x [0, d d 2 ) 2 : (x; d d 2 ) = and d i q i (x) (i =, 2)}. One would exect to be able to estimate the size of the set Λ d R Ψ by vol(r)ρ(d)(d d 2 D) 2, and indeed, we shall rove the following result: Theorem 2. (Level of Distribution). Let q i (x, y) := a i x 2 + 2b i xy + c i y 2, for i =, 2 be a air of irreducible quadratic forms in Z[X, Y ], with a i, b i, c i Z, such that Res(q, q 2 ) 0. Defining Λ d and Ψ as in () and (2), let T (M, Q) := su #(Λ d R Ψ) vol(r)ρ(d, d 2 ). d i Q i (d i ;D)= (R) M (d d 2 D) 2 Then there exist absolute constants ν and ν 2, both at least, such that T (M, Q) Q Q 2 (log 2Q Q 2 ) ν + M Q Q 2 (log 2Q Q 2 ) ν 2. We aroach this Theorem by first examining the functions ρ and ρ, then develoing uer bounds and formulæ relating the two functions. As in Daniel s work, we shall reformulate the starred roblem in terms of lattices, and use a ointcounting argument to generate the main term. The evaluation of the error term will be elementary but technical. 3
4 2. Transition from Λ d to Λ d We begin with the following bridging result which will be emloyed in Section 2.4 to exress the unstarred sum in terms of the starred sum, leading to Theorem 2.. Lemma 2. (Transition Formula). Let D N and suose that (d ; D) = (d 2 ; D) =. Then we have # (Λ d R Ψ) = # (Λ c R/b Ψ b ), b ψ(d) where c i := d i /(d i ; b 2 ), for i =, 2, the multilicative function ψ is defined by: ψ( α, β ) := max(α,β) 2, and the lattice coset Ψ b is defined by Ψ b := {x Z 2 : x b z (mod D)}, where b denotes the multilicative inverse of b modulo D. By definition, #(Λ d R Ψ) = #{x R : d i q i (x), i =, 2, x Ψ}. We artition this set according to (x; ψ(d)) to give #{x R : q i (x) 0 (mod d i ), (x; ψ(d)) = b, x Ψ}. b ψ(d) Using the easily checked facts that ψ(d)/b = ψ(c), and that (y; ψ(c)) = iff (y; c c 2 ) =, we rewrite this as #{y R/b : q i (y) 0 (mod c i ), (y; c c 2 ) =, y Ψ b }, b ψ(d) and hence the result. 2.2 Uer Bounds for ρ It is crucial to understand the number of simultaneous zeros of our quadratic forms to given moduli. Two simlifications are useful to consider. First, we consider a starred function ρ which counts the number of solutions which are corime to the modulus. Second, we consider only one form at a time. The latter simlification is similar to a roblem considered by Daniel [3]. To begin with, we note the fact that ρ and ρ are multilicative functions, reducing the roblem to evaluating the functions for rime ower arguments. 4
5 2.2. The OneForm Problem In the derivation of our uer bound for the twoform ρ function, we will be able to reduce to the simler oneform roblem. It is ossible to derive stronger results if we restrict to one form, including an elegant formula for ρ( α ), which will be utilised in Section 3. Let q be an irreducible quadratic form in the variables x and x 2. Define: ρ (a) := #{x [0, a) 2 : q(x) 0 (mod a), (x; a) = }, ρ(a) := #{x [0, a) 2 : q(x) 0 (mod a)}. Lemma 2.2. We have the following uer bounds: ρ ( α ) φ( α ) and ρ( α ) α α for all rimes and for all ositive integers α. Moreover, we have: ρ ( α ) 2φ( α ) and ρ() 2 for all ositive integers α and for all rimes which satisfy (e e 2 δ; ) =, where e and e 2 are the coefficients of the monomials x 2 and x 2 2 resectively, and δ is the discriminant of q. The argument resented below closely follows the work of Daniel [3]. Define: τ (a) := #{x [0, a) : q(x, ) 0 (mod a)}, τ 2 (a) := #{x [0, a) : q(, x) 0 (mod a)}. Note that e e 2 0, by irreducibility of the form q. Let be a rime such that does not divide e e 2. Suose x is counted by ρ ( α ) for α, that is q(x, x 2 ) 0 (mod α ) and (x; ) =. We ll show is corime to both x and x 2. Suose, for a contradiction, that x 2, then e x 2 0 (mod ), but does not divide e, hence x, a contradiction. We get a similar contradiction if we assume x. Hence the vectors x counted by ρ ( α ) are recisely those for which q(x, x 2 ) 0 (mod α ) and x, x 2 are both corime to. From this, we may deduce that (4) ρ ( α ) = τ ( α )φ( α ) = τ 2 ( α )φ( α ) for α, if does not divide e e 2. To derive the second identity, for examle, note that ρ ( α ) = #{x [0, α ) 2 : q(x) 0 (mod α ), (x ; ) = (x 2 ; ) = } = #{x [0, α ) 2 : q(, x 2 x ) 0 (mod α ), (x ; ) = (x 2 ; ) = } = #{y [0, α ) 2 : q(, y 2 ) 0 (mod α ), (y ; ) = (y 2 ; ) = }, 5
6 where x denotes the inverse of x modulo α. Even if e e 2, we know that does not divide x or does not divide x 2, so ρ ( α ) (τ ( α ) + τ 2 ( α ))φ( α ) for all and α. We now aly the following Theorem of Huxley [7]: Theorem 2.2. If g Z[X] is a olynomial of degree n 2 and nonzero discriminant δ, then for all rime owers e, t( e ) n m/2, where m δ and where t(a) := #{x [0, a) : g(x) 0 (mod a)}. The olynomials q(x, ) and q(, x) are irreducible hence have no reeated root; so the Theorem alies, giving τ ( α ) + τ 2 ( α ) 4δ /2 uniformly for all and α. Hence ρ ( α ) φ( α ). Huxley s Theorem also alies for the secial case where (e e 2 δ; ) =. As is corime to the discriminant, we have that τ ( α ) 2 for all α, and an alication of (4) gives ρ ( α ) 2φ( α ). We can exress ρ in terms of ρ, and we have (5) ρ( α ) = 2(α α/2 ) + 2β ρ ( α 2β ). 0 β< α/2 To see this, we emloy the more general result, equation (7), that ρ(d) = b ψ(d) ( ) ρ (d ; b 2 )(d 2 ; b 2 2 ) (c), b for a twoform function ρ(d), where ψ( α, β ) = max(α,β)/2, and c i = d i /(d i ; b 2 ). This immediately gives the desired result by reducing to the oneform function ρ(d) := ρ(d, ). Formula (5) gives our bound for ρ, and Lemma 2.2 is roved. For certain alications, we shall need a more exact formula, which is rovided by the following Lemma: Lemma 2.3. Let q(x, x 2 ) = ax 2 + 2bx x 2 + cx 2 2 be an irreducible quadratic form. Suose that does not divide 2acδ, then ( ) δ ρ( α ) = φ( α ){ + } α/2 + 2(α α/2 ), where δ := b 2 ac is the discriminant of q. 6
7 By the argument of Lemma 2.2, we have that ρ ( α ) = φ( α )τ ( α ), where τ ( α ) is the number of solutions of q(x, ) 0 (mod α ). Now x is such a solution if and only if (x + a b) 2 a (a b 2 c) (mod α ), where a is the multilicative inverse of a modulo α. By Hensel s Lemma, we only need count the number of solutions to this equation modulo, so ( ) ( ) a τ ( α (a b 2 c) δ ) = τ () = + = +, hence ρ ( α ) = φ( α ){ + ( δ ) }. Alying equation (5), the result follows The TwoForm Problem We now consider the twoform variants of ρ and ρ, defined in (3) The Function ρ Lemma 2.4. For every rime one has that ρ ( e, f ) max(e,f), and if is corime to Res(q, q 2 ), then: (6) ρ ( e, f ) = 0, if e > 0 and f > 0. First, we shall show that for all, ρ ( e, f ) max(e,f). We may assume that ρ ( e, f ) 0. Thus there exists x = (x, x 2 ) such that (x; ) =, q (x) 0 (mod e ), and q 2 (x) 0 (mod f ). Without loss of generality, assume does not divide x 2. Define Q i (Y ) := q i (Y, ), and y x x 2 (mod max(e,f) ), where x 2 is the multilicative inverse of x 2 modulo max(e,f). Now 0 q (x, x 2 ) x 2 2Q i (y) (mod e ), and does not divide x 2, hence e Q (y). Similarly, f Q 2 (y), whence min(e,f) Q (y) and min(e,f) Q 2 (y). We deduce that min(e,f) Res(q, q 2 ). This restricts what min(e, f) can be. Precisely, define m by m Res(q, q 2 ), then min(e, f) m. Note that for fixed q, q 2, we have m = 0 for all but finitely many. Now ρ ( e, f ) #{x (mod e+f ) : f q 2 (x), (x; ) = } = 2e #{x (mod f ) : f q 2 (x), (x; ) = }, whence ρ ( e, f ) 2e f, where we have used the oneform result, Lemma 2.2. A similar argument gives ρ ( e, f ) 2f e, which combine to give: ρ ( e, f ) 2m max(e,f). 7
8 Now m = 0 for all but finitely many, so ρ ( e, f ) max(e,f). Moreover, we extract from the roof that if (; Res(q, q 2 )) =, and if min(e, f) > 0, then min(e, f) > m, and hence ρ ( e, f ) = 0, which imlies the result (6), and our Lemma is roved The Function ρ Lemma 2.5. For every rime and for all nonnegative integers e and f, let m := min(e, f), M := max(e, f), then ρ( e, f ) (M m + ) 2m+M and ρ(, ) 2. Moreover, for all but a finite set of rimes, one has ρ(, ) 2 and ρ(, ) 2. Assume e = min(e, f). Recall the definitions () of Λ d and Λ d. We may write ρ(d) = #(Λ d (0, d d 2 ] 2 ) and ρ (d) = #(Λ d (0, d d 2 ] 2 ). Alying Lemma 2. and using the notation of Section 2., we have: (7) ρ(d) = #(Λ c (0, d d 2 /b] 2 ) = ( ) ρ (d ; b 2 )(d 2 ; b 2 2 ) (c), b b ψ(d) where c i = d i /(d i ; b 2 ). In articular, (8) ρ( e, f ) = 0 β f/2 b ψ(d) ( ) ( ) ρ e ( e ; 2β ), f ( e ; 2β )( f ; 2β 2 ). ( f ; 2β ) β Slit the range of summation as 0 2β e, e < 2β < f, and β = f/2. We have the following uer bound for ρ( e, f ): ρ ( e 2β, f 2β ) 6β + ρ (, f 2β ) 2e+2β + ( e+f f/2 ) 2 0 β e/2 0 β e/2 f+4β + e/2<β<f/2 (f e) 2e+f + 2e+f, e/2<β<f/2 2e+f + 2e+f as required. We can do a little better in secial cases. When e = 0 or f = 0, ρ( e, f ) reduces to the oneform roblem, so by Lemma 2.2 we have ρ(, ) 2, ρ(, ) 2, for all but finitely many rimes. Also, by equation (8), we have (9) ρ(, ) = ρ (, ) + 2 = O( 2 ), comleting the roof of Lemma
9 2.3 Level of Distribution Starred Version In the calculation of our sum, we need to evaluate #(Λ d R Ψ). Here, Ψ is defined to be the lattice coset {x Z 2 : x z (mod D)} for chosen z and D which deend only on the forms in question. We will see that it is only necessary to consider those d for which (d ; D) = (d 2 ; D) =. The level of distribution formula gives us the error term involved in estimating #(Λ d R Ψ) by vol(r)ρ(d)/(d d 2 D) 2, as we average over d and R. As mentioned in the background section, it is simler to deal first with a starred level of distribution formula. This is one in which we imose corimality conditions. We have: Lemma 2.6. Define T (M, Q) := su R: R M d i Q i (d i ;D)= #(Λ d R Ψ) ρ (d, d 2 ) (d d 2 D) vol(r) 2. If q i (x, y) = a i x 2 +2b i xy+c i y 2 (for i =, 2) are a air of irreducible quadratic forms in Z[X, Y ] such that a i, b i, c i Z and such that Res(q, q 2 ) 0, then one has T (M, Q) M Q Q 2 (log(2q Q 2 )) Q Q 2 (log(2q Q 2 )) 6 uniformly for M > 0 and Q, Q The Quantities Λ d Assume that d = (d, d 2 ) is fixed and define a := d d 2. Let U(a) be the set of equivalence classes of x Z 2 under multilication with (x ; x 2 ; a) =. That is, if (x; a) = (y; a) =, then we define a relation by: x y iff λ Z, x λy (mod a). Our motivation for this definition is that we wish to artition Λ d into equivalence classes. It is easily checked that defines an equivalence relation, and that (λ; a) =. Moreover, suose y A for some A U(a), then if (λ; a) =, we have λy A. In fact, for a fixed y A, A = {x Z 2 : x λy for some λ with (λ; a) = }. Bringing the forms into lay, one may verify that if y Λ d, then λy Λ d, given (λ; a) =. So for a given A U(a), either A Λ d or A Λ d =. 9
10 This suggests the definition U (d) := {A U(d d 2 ) : A Λ d }. Hence we may artition Λ d into disjoint sets as follows: Λ d = A. A U (d) For a given A, fix y A. Then for any x A, we have that x λy (mod a), for (λ; a) =. The vector x is uniquely determined modulo a by λ, so there are exactly φ(a) choices for x modulo a. Stated another way, #(A [0, a) 2 ) = φ(a). Hence, (0) ρ (d, d 2 ) = #U (d)φ(d d 2 ). Return to our summand: #(Λ d R Ψ) ρ (d, d 2 ) (d d 2 D) vol(r) 2 = #(A R Ψ) ρ (d, d 2 ) (d d 2 D) vol(r) 2 A U (d) () { vol(r)} = ρ (d, d 2 ) #(A R Ψ) #U (d)(d d 2 D) 2 A U (d) #(A R Ψ) φ(d d 2 ) (d d 2 D) vol(r) 2, A U (d) where we have used equation (0) in the last line Estimating #(A R Ψ) The next task is to calculate the quantity #(A R Ψ). Our line of attack will be to introduce lattices G(A) generated by the sets A. Alying techniques from the geometry of numbers, we will exress the error in terms of a minimal basis of G(A). Choose A U(a) and define G(A) by: G(A) := {x Z 2 : ( λ Z)( y A)(x λy (mod a))}. Fix y 0 A, then we may rewrite G(A) as: G(A) = {x Z 2 : ( λ Z)(x λy 0 (mod a))}. 0
11 and it becomes clear that G(A) is the sublattice of Z 2 generated by the vectors of A. We can see that A = {x G(A) : (x; a) = }, so #(A R Ψ) = µ(b) x b (x;a) G(A) R Ψ = µ(b) #{x R/b : bx G(A) Ψ}, b a The aearance of bx G(A) in our equation for #(A R Ψ) suggests that we should work modulo a/b, and motivates the following definition: given c a, and given A U(a), we define A (mod c) to be unique element of U(c) such that A A (mod c). If b a, then bx G(A) iff x G(A (mod a/b)). Moreover, bx Ψ iff x b z (mod D). The inverse exists as b a = d d 2, and we ve assumed that each d i is corime to D. Define Ψ := {x Z 2 : x b z (mod D)}, then (2) #(A R Ψ) = b a µ(b) #(R/b G(A (mod a/b)) Ψ ). Define R := R/b, a := a/b, A = A (mod a ). Then, bearing in mind that det(g(a )) = a, an alication of Lemma 2. in [3] allows us to deduce: Lemma 2.7. There exist vectors v (), v (2) G(A ) satisfying the following roerties:. the air (v (), v (2) ) is a basis of G(A ), 2. v () = min{ v : v G(A ) \ {0}}, and 3. a v () v (2) 2 3 a. Let θ : R 2 R 2 be the automorhism which mas the canonical basis of R 2 to (v (), v (2) ). Then θ has matrix ( ) v () v (2) v () 2 v (2), 2 and det θ = [Z 2 : G(A )] = a. We ll now derive a simle condition for x G(A ) Ψ. First note that although b z may not be a member of G(A ), we may find a reresentative z of b z modulo D which is in G(A ), as D is corime to a. Write x =
12 θa, z = θb. As (D; a ) =, we may invert θ modulo D and deduce that x z (mod D) iff a b (mod D). Hence x R G(A ) Ψ iff a θ (R ) Z 2 and a b (mod D). Write a = b + Dc. The above condition is equivalent to c Z 2 { D (θ (R ) b)}. We are now in a osition to be able to estimate the number of lattice oints on D (θ (R ) b). The error term in aroximating the number of lattice oints enclosed by a curve C by its area vol(c) is given by O( (C) + ). The reader is referred to Lemma 2.. in [8] for further details. An alication of this result gives: (3) #(R G(A ) Ψ ) = D 2 vol(θ (R )) + O( (θ (R )) + ) and (4) vol(θ (R )) = vol(r ) = b a a b vol(r) = 2 ab vol(r). ( ) Note θ = v (2) 2 v (2) a v () 2 v (), hence θ θ (u) := max u R 2 \0 u v(2) a v (), a max( v (), v (2) ) by Lemma 2.7. Also, (θ (R )) θ (R ) and (R ) = (R)/b, so (θ (R )) (R) b v () M b v (). Substituting this and equation (4) into (3) gives: (5) #(R G(A ) Ψ ) = vol(r) ( ) M + O abd 2 b v () +. In the sequel, we shall write v(a ) for v () in order to secify the equivalence class. 2
13 We know v(a ) λy (mod a/b) for some λ Z and some y A. Then bv(a ) (λb)y (mod a), so bv(a ) G(A), hence b v(a ) v(a). Insert this into equation (5): #(R/b G(A (mod a/b)) Ψ ) = vol(r) abd 2 ( ) M + O v(a) +. Then substituting this into equation (2) gives: #(A R Ψ) = ( ( )) vol(r) M µ(b) + O abd 2 v(a) + b a = φ(a) ( ( )) M (6) a 2 D vol(r) + O d(a) 2 v(a) +. Finally, substituting this into equation (), we have: T (M, Q) su #(A R Ψ) vol(r)φ(d d 2 ) (d d 2 D) 2 (7) d i Q i A U R: R M (d) (d i ;D)= d i Q i A U (d) (d i ;D)= M Evaluating T (Q) d i Q i d(d d 2 ) ( ) M d(d d 2 ) v(a) + A U (d) = MT (Q) + T 2 (Q), say. It is our aim in this section to rove: Lemma 2.8. The quantity satisfies the uer bound: T (Q) := d i Q i d(d d 2 ) v(a) + A U (d) d i Q i d(d d 2 )#U (d) v(a) T (Q) Q Q 2 (log 2Q Q 2 ) For A U (d), we have v(a) d d 2, by Lemma 2.7, and there exist y A, λ Z such that v(a) λy (mod d d 2 ). Therefore, for i =, 2 : 3
14 q i (v(a)) λ 2 q i (y) 0 (mod d i ), so d i q i (v(a)). Consequently, T (Q) 0< v Q Q 2 v d i Q i d i q i (v) d(d d 2 )#U (d). As one would exect, we shall roceed by tackling the innermost quantity first. To begin with, we shall relace d(d d 2 )#U (d) with a simler multilicative function. The first ste is the roof that #U (d) 2 ν(d d 2 ). Write d = e, and d 2 = f and aly equation (0): #U (d) = ρ (d) φ(d d 2 ) = ρ ( e, f ) φ( e+f ), by multilicativity of ρ. For the rest of this section, define P := Res(q, q 2 ). By Lemma 2.4, if (; P ) = and if min(e, f) > 0, then ρ ( e, f ) = 0. Thus #U (d) = 0 unless for all satisfying (; P ) =, we have min(e, f) = 0. In which case, we may write: #U (d) : (;P )= : (;P )= ρ ( e, ) φ( e ) ρ ( e, ) φ( e ) : (;P )= : (;P )= ρ (, f ) φ( f ) ρ (, f ) φ( f ). : P C max(e,f) φ( e+f ) Now define ρ (b) := ρ (b, ) and ρ 2(b) := ρ (, b). By an alication of Lemma 2.2, there exist integers P and P 2, deending only on the forms q and q 2, such that ρ i ( α ) 2φ( α ) if (; P i ) =, and ρ i ( α ) α for all α. Hence, #U (d) : (;P P )= : 2 (;P P )= 2 ν(d d 2 ), ρ ( e ) φ( e ) : (;P P 2 )= : (;P P 2 )= 2 C : P ρ 2( f ) φ( f ) C : P 2 C : P ρ ( e ) φ( e ) C : P 2 ρ 2( f ) φ( f ) as required, bearing in mind that each factor of 2 comes from a rime which either divides d or d 2, but not both. 4
15 Observing that 2 ν(d d 2 ) d(d d 2 ), and using the submultilicativity roerty of d, we deduce that whence T (Q) d(d d 2 )#U (d) d(d ) 2 d(d 2 ) 2, 0< v Q Q 2 v d i q i (v) d(d ) 2 d(d 2 ) 2. Defining the function h by h(n) := a n d(a)2, we have (8) T (Q) 0< v Q Q 2 v h(q (v))h(q 2 (v)) The function h Our aroach to the evaluation of this sum will be to decomose it into dyadic intervals. Unfortunately, the arguments of the function h are q (v) and q 2 (v) which are, in order of magnitude, the square of our summation variable v. To facilitate the decomosition, we shall emloy Lemma 2.0 below, which guarantees the existence of a divisor m i of q i (v) of the correct order of magnitude, such that we can relace h(q i (v)) with h(m i ) 5. To verify the conditions of Lemma 2.0, we shall need the following reliminary Lemma: Lemma 2.9. The function h is multilicative. Moreover, h is submultilicative in the sense that for all m, m 2, we have h(m m 2 ) h(m )h(m 2 ). Furthermore, we have h() uniformly in. Multilicativity is trivial. To rove submultilicativity, it suffices to show that h( e+f ) h( e )h( f ). Now e+ h( e ) = i 2 = (e + )(e + 2)(2e + 3)/6, i= giving h( e )h( f ) h( e+f ) = ef(6 + 8f + 8e + 26e 2 + 8e 2 f + 8ef + 4e 2 f 2 + 8ef f 2 ), this is clearly nonnegative, demonstrating submultilicativity. We also have h() = 5 uniformly for all, comleting the roof of Lemma 2.9. This is sufficient to satisfy the conditions of the following Lemma, which is to found in [3] as Lemma 2.2: 5
16 Lemma 2.0. Let h be some ositive submultilicative arithmetical function such that h() uniformly in. Let η, then for every natural number n, there exists a ositive integer m satisfying m n, m n /η, and h(n) η h(m) + η. We aly Lemma 2.0 to equation (8), with η = 4, to obtain: T (Q) j 0 P =2 j Q Q 2 P P v 2P m i q i (v) m i P /2 i=,2 h(m ) 5 h(m 2 ) 5, where we have slit the range for v into dyadic intervals, and used the fact that q i (v) v 2 to deduce that m i q i (v) /4 imlies m i P /2. We have T (Q) j 0 P =2 j Q Q 2 P m,m 2 P /2 h(m ) 5 h(m 2 ) 5 v 2P q i (v) 0 (mod m i ) i=,2 The innermost sum is of order ρ(m, m 2 ){P 2 /(m m 2 ) 2 + P/(m m 2 )}, where the second term accounts for the error at the boundary. We have arranged that m m 2 P, whence the second term is subsumed by the first, and hence the innermost sum is of order ρ(m, m 2 )P 2 /(m m 2 ) 2, so T (Q) P j 0 P =2 j Q Q 2 h(m )5 h(m 2 )5 ρ(m, m 2 ) m 2 m,m 2 m 2 P /2 2 h(m Q Q )5 h(m 2 )5 ρ(m, m 2 ) 2. m 2 m,m 2 (Q Q 2 m 2 ) /4 2 At this oint, we marshal together the facts we have uncovered concerning the function ρ. We know that ρ is multilicative and we have the results of Lemma 2.5, including ρ(, ) 2. We have that ρ( e, f ) (M m + ) 2m+M, where m = min(e, f) and M = max(e, f), and, by the oneform roblem, ρ( e, ), ρ(, e ) e e. Moreover, there exists a natural number P such that if (; P ) = then ρ(, ) 2 and ρ(, ) 2. Note that for a doubly multilicative function g, one has m,m 2 Q g(m, m 2 ) Q g( e, f ). e,f=0. 6
17 Alying this to the above exression for T, we deduce T (Q) Q Q 2 (Q Q 2 ) /4 e,f=0 h( e ) 5 h( f ) 5 ρ( e, f ) 2e+2f. Let g( e, f ) denote the summand. In order to be able to aly our uer bound for ρ, we slit the sum as follows: g( e, f ) = f=0 e=0 f e g( e, f ) + g( e, f ) =: S + S 2. f=0 e=0 e= f=0 Let be a good rime, in the sense that (; P ) =, then, bearing in mind that ρ(, ) 2, we have: S C 2 + C f f=2 e=0 f 6 e 5 f. Call the double sum S, then 2 S f=2 f 32 f 32 f 2 2, f 2 f=2 and this is convergent by the ratio test, so S = O(/ 2 ), hence S / + C / 2. Similarly, S / + C/ 2. So S + S C 2, given that is any good rime. The analysis above shows that S + S 2 < even for the finitely many bad rimes such that P. Thus we have: T (Q) Q Q 2 ( C ), 2 (Q Q 2 ) /4 and an alication of Mertens Theorem leads to our uer bound T (Q) Q Q 2 (log 2Q Q 2 ) To be recise, we have used the following: 7
18 Lemma 2.. Let Q > and C > 0 be real numbers. Let k be a natural number and define S = ( + k + C ). 2 Q Then Observe S = Q S k,c (log Q) k. ( ) k ( ) k ( + k + C ). 2 Q Define a function f : [0, ] R by f(x) = ( x) k ( + kx + Cx 2 ), then f(x) = ( kx ±x k )( + kx + Cx 2 ), so there exist constants c i deending on k such that: f(x) = + c 2 x c k+2 x k+2 + c 2 x c k+2 x k+2 + { c c k+2 }x 2 + Lx 2 ( + x 2 ) L, where L deends only on k and C. Recall Mertens Theorem, which states ( ) ( ) = e γ log z + O. (log z) 2 This gives us Now S Q z ( ) k ( + ) L (log Q) ( k + ) L. 2 2 Q Q ( + ) = 2 Q n = Q so S (log Q) k (C ) L k,c (log Q) k, as required Evaluating T 2 (Q) We shall rove n =: 2 C n2 n= Lemma 2.2. The quantity T 2 satisfies the uer bound T 2 (Q) Q Q 2 log(2q Q 2 ) 6. 8
19 Recall T 2 (Q) := d(d d 2 )#U (d). d i Q i i=,2 In our analysis of the sum T (Q), we demonstrated that #U (d) 2 ν(d d 2 ). We have that 2 ν(a) d(a) for any a and that the d function satisfies d(ab) d(a)d(b) for any a and b. Thus, ( ) ( ) T2 (Q) d(d ) 2 d(d 2 ) 2 Q Q 2 (log Q ) 3 (log Q 2 ) 3 d Q d 2 Q 2 Q Q 2 (log 2Q Q 2 ) 6. In the last line, we use the AMGM inequality to deduce (9) (log A ) n (log A 2 ) n (log A A 2 ) 2n This roves Lemma 2.2. Combining this with Lemma 2.8 gives us our starred level of distribution formula, Lemma Level of Distribution Unstarred Version Recall our convention that the symbol c i reresents d i /(d i ; b 2 ). We aly Lemma 2. and equation (7) to give the following exression for T (M, Q): { vol(r/b)} su #(Λ c R/b Ψ b ) ρ (c) R: (c c 2 D) 2 d i Q i (R) M (d i ;D)= c i Q i b Q Q 2 (c i ;D)= (b i ;D)= b ψ(d,d 2 ) δ(q, c, b) su R: (R) M #(Λ c R/b Ψ b ) ρ (c) (c c 2 D) vol(r/b) 2, where δ(q, c, b) = #{(d, d 2 ) : d i Q i, c i = d i /(d i ; b 2 ), b ψ(d, d 2 )}. We shall derive an uer bound for δ. The aroach used is to fix a rime and to consider quantities β, α i, and γ i for i =, 2 such that α i d i, γ i c i, and β b. We take the quantities β and γ i to be fixed: our task is to count the number of ossibilities for α i. For a fixed rime, it may be verified that there are at most 8β ossibilities, whence δ(q, c, b) β b 8β =: g(b). By induction on β, we have that 8β ( ) β+7 7, thus emloying the fact that d 8 (b) = ( β+7 ) β b 7, we have g(b) d8 (b). Consequently, d 8 (B) B(log B) 7, b B δ(q, c, b) b B g(b) b B 9
20 where the last ste uses b B d k(b) B(log B) k, which follows from (2..4) in [0]. Note that we are summing over c i Q i, b Q Q 2, but we may restrict the range of summation by observing a relationshi which holds when δ(q, c, b) 0. Suose that δ(q, c, b) 0, then there exist d, d 2 such that b ψ(d, d 2 ) and c i = d i /(d i ; b 2 ). It is easily verified that b ψ(d, d 2 ) imlies b d d 2, and hence b (d ; b 2 )(d 2 ; b 2 ). This may be rewritten as c c 2 b d 2 d 2, from which it follows that c c 2 b Q Q 2. For the sake of simlicity, we shall relace the exression #(Λ c R/b Ψ b ) ρ (c) (c c 2 D) vol(r/b) 2 with L(c, b, R). Then our sum T (M, Q) is estimated by T (M, Q) δ(q, c, b) L(c, b, R) c,c 2 : (c i ;D)= c i Q i j i : C i =2 j i Q i b: (b;d)= c c 2 b Q Q 2 C i c i 2C i (c i ;D)= b: (b;d)= b Q Q 2 c c 2 su R: (R) M d 8 (b) su R: (R) M L(c, b, R). If we further slit the range for b into dyadic intervals, then T (M, Q) d 8 (b) su L(c, b, R). j i : (R) M C i =2 j i Q i C i c i 2C i (c i ;D)= k: B=2 k Q Q 2 c c 2 b: (b;d)= B b 2B Our aim is to use the estimate for b d 8(b), but we need to handle sensitively the factor of su L(c, b, R). For each choice of B, define b(b) by requiring B b(b) 2B, (b(b); D) = and requiring that for all b with B b 2B and (b; D) =, one has su L(c, b, R) su L(c, b(b), R). (R) M (R) M Let S denote the set of integers B such that there are no b in the range 20
21 B b 2B with (b; D) =. We have the uer bound: T (M, Q) B(log 2B) 7 su j i : C i =2 j i Q i j i : C i c i 2C i (c i ;D)= k: C i =2 j i Q i B=2 k Q Q 2 C C 2 B S k: B=2 k Q Q 2 c c 2 B S B(log 2B) 7 C i c i 2C i (c i ;D)= (R) M su (R) M L(c, b(b), R) L(c, b(b), R). Writing R := R/b(B), we may now aly our starred level of distribution formula (Lemma 2.6) to the inner sum, which is bounded from above by su #(Λ c R Ψ b(b) ) ρ (c) (c c 2 D) 2 vol(r ) so C i c i 2C i (c i ;D)= (R ) M/B M B C C 2 (log 8C C 2 ) C C 2 (log 8C C 2 ) 6, T (M, Q) M j i : C i =2 j i Q i C C 2 (log 8C C 2 ) C C 2 (log 8C C 2 ) j i : C i =2 j i Q i 2 Q k log Q 2 2 C C 2 (log 2 Q k log Q 2 2 C C 2 k (log 2 k+ ) 7. Estimating the sums over k by the aroriate integrals, we arrive at T (M, Q) M j i : C i =2 j i Q i C C 2 (log 8C C 2 ) (log 2Q Q 2 ) C C 2 (log 8C C 2 ) Q Q 2 (log Q Q 2 ) 7 C j i : C 2 C i =2 j i Q i M(log 2Q Q 2 ) 8 2 (j+j2)/2 (log 2 j+j2+3 ) 2255 j i log 2 Q i + Q Q 2 (log 2Q Q 2 ) 7 (log 2 j+j2+3 ) 6. j i log 2 Q i Once more we estimate the sums via integrals to give T (M, Q) M(log 2Q Q 2 ) 8 Q Q 2 (log 2Q ) (log 2Q 2 ) Q Q 2 (log 2Q Q 2 ) 7 (log 2Q ) 7 (log 2Q 2 ) 7. k+ ) 7 2
22 Finally, we bring the result into the desired form by alying the AMGM inequality as in equation (9). This roves the level of distribution formula. 3 Pairs of Forms with Almost Prime Values In the next section, we set the scene by introducing the terminology of sieves, before going on to the derivation of Theorem. in Section The Terminology of Sieves Sieve methods aim at finding the rimes in a multiset (essentially a sequence) of natural numbers A. Tyically, one defines a sifting set P of rimes then one tries to discover the value of the sifting function: S(A, P, z) := {a : a A, if P and a, then z}. This is useful in giving bounds for the number of rimes in A. In our case, we are examining almostrimes, so we will want a lower bound for {P 5 : P 5 A}, where our multiset A will be A := {q (x, y)q 2 (x, y) : (x, y) Z 2 XR (0) Ψ}, and, taking D = 6 Res(q, q 2 )a a 2 c c 2 δ δ 2, as in the statement of Theorem.,we define Ψ := {x Z 2 : x z (mod D)}, where z is chosen such that (q (z); D) = (q 2 (z); D) =. For the sifting set P, we shall take all rimes which do not divide D. In the evaluation of the sifting function, it is necessary to consider a number of auxiliary quantities, including A d := {a : a A, a 0 (mod d)}. We will need to use an aroximation Y for the number of elements in the set A. In the case under consideration, it is natural to take Y = X 2 vol(r (0) )/D 2. We shall choose the function ω() such that Y ω() = { A aroximately, for P, 0, for P, where P is the comlement of P in the set of all rimes. We extend the definition of ω by multilicativity to all squarefree numbers. The quantity R d is, in some sense, the error in aroximating A d by Y ω(d) d, that is, we define: R d := A d ω(d) Y, if µ(d) 0. d 22
23 3.2 Proof of the Main Result Our main tool will be the following weighted sieve of Diamond and Halberstam [5]: Theorem 3.. With the notation of Section 3., suose there exist real constants κ >, A, A 2 2, and A 3 such that (A) 0 ω() <, (B) (C) z <z ( ω() ) ( ) κ ( log z + A ), 2 z < z, log z log z d<y α /(log Y ) A 3 (d;p)= µ 2 (d)4 ν(d) R d A 2 Y log κ+ Y, for some α with 0 < α ; that (D) (a; P) = for all a A, and that (E) a Y αµ for some µ, and for all a A. There exists a real constant β κ > 2 such that for any real numbers u and v satisfying α < u < v, β κ < αv, we have whenever {P r : P r A} Y (20) r > αµu + κ f κ (αv) v/u <Y /v ( ω() ), ( F κ (αv s) u ) ds v s, where f k and F k are solutions to a system of delay differential equations secified in [5]. Let I(κ, α, µ) denote the minimum value of the lower bound on the right of (20) as u and v vary, subject to the above constraints. As tabulated in [5], I(2,, 2) < 5. 23
24 An examination of Diamond and Halberstam s aer shows that I is a continuous function of α and µ, so for our uroses, it will be sufficient to demonstrate that Theorem 3. alies for κ = 2, and for any α <, and µ > Condition (A) We shall now verify the conditions required for the alication of Theorem (3.), critically emloying the level of distribution formula in the estimation of the errorsum (C). To begin, we need to formulate an aroriate definition for the quantity ω(). Recall that for P, we would like Y ω()/ to be roughly A, so we need an estimate for A. Writing Ω = XR (0) Ψ, we have: A = #{(a, b) Ω : q (a, b)q 2 (a, b)} = #{(a, b) Ω : q (a, b)} + #{(a, b) Ω : q 2 (a, b)} #{(a, b) Ω : q (a, b), q 2 (a, b)} = #(Λ (,) Ω) + #(Λ (,) Ω) #(Λ (,) Ω). Now, we have the aroximation: whence #(Λ d R Ψ) ρ(d, d 2 ) (d d 2 D) 2 vol(r), A X2 vol(r (0) ) {ρ(, ) + ρ(, )} X2 vol(r (0) ) ρ(, ). 2 D 2 4 D 2 This leads us to define { (ρ(, ) + ρ(, )) 3 ρ(, ) P ω() = 0 P. With this definition, we may quickly verify condition (A). First, we must check that 0 ω(). We may assume that P, and by equation (9), we have ρ(, ) = ρ (, ) + 2. As (; D) =, Lemma 2.4 rovides us with ρ (, ) = 0, so ω() = (ρ(, ) + ρ(, ) ), but, from the definition, ρ(, ), so ω() 0. On the other hand, by Lemma 2.3, we have ( ( δ ρ(, ) = + ( ) + )), ρ(, ) = + ( ) 24 ( + ( )) δ2,
25 so, writing χ i () := ( δ i ), ω() = 2 + χ () + χ 2 () ( + χ () + χ 2 ())/, whence ω() 4 <, as we ve assumed 5. Incidentally, this inequality exlains the factor of 6 in our choice of D Condition (B) This condition exresses the κdimensionality of the sieve roblem. One should think of the quantity ω()/ as being the robability that an element of A is divisible by, and that κ is the average value of ω(), in some sense. In many sieve roblems, one finds that κ =, a linear sieve. However, in our roblem, we will demonstrate that κ = 2, as one would exect from the above definition of ω(). We must rove z <z ( ω() ) ( ) κ ( log z + A ), 2 z < z. log z log z Without loss of generality, we may assume that z 5, as ω() = 0 if = 2 or 3. So uon taking logs, we must demonstrate z <z i= ω() i i i κ log log z κ log log z + log( + A / log z ), for z 5. Now for any B > 0 there exists a constant A such that B x log(+a x) whenever 0 x, so we may relace log(+a / log z ) in the above equation by B / log z. We exect the sum z <z ω()/ to contribute the main term, and begin by considering the error term, bearing in mind that ω() 4 for all rimes. as required. ω() i 4i i i in i i=2 z <z i=2 n z ( 4 i /i x=z x dx + ) /z i / log z, z i=2 25
26 The main term is z <z ω()/, which exands to: z <z 2 + z <z χ () + z <z = 2 log log z 2 log log z + z <z χ 2 () χ () z <z + z <z + χ () + χ 2 () 2 χ 2 () + O(/z ). In estimating the sums involving characters, we use a result of Mertens, to be found in Chater 7 of [4], that for any nonrincial character χ, one has χ() log = O(). So z z χ() = z z χ() log log z z χ() log. log z log z This comletes our verification of condition (B). We see that κ, the dimension of the sieve, has the value κ = Condition (C) Condition (C) is concerned with the quantity R d := A d ω(d) d Y, for squarefree d. Essentially, we shall sum R d as d varies in some range. In this roblem, the range of summation is referred to as the level of distribution, and it is our aim is to ensure that the level of distribution is as large as ossible, whilst requiring that the sum be bounded above by Y/(log Y ) 3. We would like to bring to bear our work on the level of distribution formula, and thus to relate A d to quantities of the form #(Λ c XR (0) Ψ). Our goal is fulfilled by the following formula: Lemma 3.. A d = µ c,c 2 d d c c 2 ( c c ) 2 #(Λ c XR (0) Ψ). d 26
27 For the duration of this roof, let us write Ω for Z 2 XR (0) Ψ, then A d = #{x Ω : (q (x); d) = d and d 2 q 2 (x)} d d 2 =d = d d 2 =d = d d 2 =d x Ω d i q i (x) (q (x)/d ;d 2 )= e d 2 µ(e) Write c = ed, and c 2 = d/d, then A d = x Ω d e q (x) d 2 q 2 (x) d c c 2 c,c 2 d µ = d d 2 =d x Ω e q (x)/d d i q i (x) e d 2 = e,d : ed d µ(e) ( c c ) 2, d x Ω c q (x) c 2 q 2 (x) x Ω ed q (x) d/d q 2 (x) µ(e) and hence the result. Naturally, it would be advantageous to exress ω(d)/d in a similar form. Indeed, we may write ω(d) d = d ω() = µ c,c 2 d d c c 2 ( c c ) 2 ρ(c, c 2 ) d (c c 2 ), 2 whence ( R d = c c ) { 2 µ #(Λ c XR (0) Ψ) Y ρ(c }, c 2 ) d (c c 2 ) 2 c,c 2 d d c c 2 #(Λ c XR (0) Ψ) Y ρ(c, c 2 ) (c c 2 ) 2. c,c 2 d d c c 2 Our ultimate aim is to derive a level of distribution of the form Y α, for any ositive α <. We consider the sum: E := µ 2 (d)4 ν(d) #(Λ c XR (0) Ψ) Y ρ(c, c 2 ), d<y α (d;p)= c,c 2 d d c c 2 27 (c c 2 ) 2
28 and we desire an uer bound for E. Let us introduce another variable, k, which secifies the highest common factor of c and c 2. Define the sets U and T k by U := {d Z : (d; P) =, µ 2 (d) = }, and This leads to the exression E = T k := {(c, c 2 ) U 2 : (c ; c 2 ) = k}. d<y α d U µ 2 (d)4 ν(d) k<y α k U (c,c 2 ) T k c,c 2 d d c c 2 c c 2 ky α..., where... := #(Λ c XR (0) Ψ) Y ρ(c,c 2 ). (c c 2 ) Note that 2 [c, c 2 ] = c c 2 /k, so the condition c, c 2 d imlies that c c 2 /k d, and hence that c c 2 dk. This is the origin of the extra condition c c 2 ky α in the inner sum. We now swa the order of summation: E k<y α k U... µ 2 (d)4 ν(d). (c,c 2 ) T k d c c 2 c c 2 ky α Consider the inner sum. We have that d m µ2 (d)4 ν(d) = 5 ν(m) ε m ε for any ositive ε. Alied to our roblem, the inner sum is bounded from above by Y ε, leading to E ε Y ε k<y α k U... (c,c 2 ) T k c c 2 ky α We will make use of the divisibility roerties of c and c 2 to examine the inner sum, which I shall denote by E(k). Write c i = kg i. Then (g ; g 2 ) =. We claim that the ma x x/k is a bijection from Λ c XR (0) Ψ to Λ g k XR (0) Ψ k. Clearly it is sufficient to rove that the given ma is a bijection from Λ c to Λ g. If x Λ c then kg i q i (x), for i =, 2. Hence q i (x) 0 (mod k), for i =, 2; but k is squarefree and corime to the resultant of q and q 2, so, by an alication of the Chinese Remainder Theorem, we must have that x 0 (mod k). Write x = ky, for some y Z 2. Another direct alication of the fact that x Λ c gives g i k 2 q i (y) for i =, 2. Now, as c i is squarefree for i =, 2, we have that (k; g i ) =, for i =, 2, so we may deduce from g i k 2 q i (y) that g i q i (y) for i =, 2, and hence that y Λ c. It is trivial to demonstrate that if y Λ g, then ky Λ c, comleting the roof of bijectivity. 28
29 To deal with the ρ term, note that ρ(c, c 2 ) (c c 2 ) 2 = ρ(kg, kg 2 ) (k 2 g g 2 ) 2 = ρ(k, k) k 4 ρ(g, g 2 ) (g g 2 ) 2, where we use multilicativity of ρ and the corimality of k and g g 2 in the last line. Recall that the only solution of q i (x) 0 (mod k) is the trivial solution x 0 (mod k). This allows us to deduce that the quantity ρ(k, k) is equal to k 2. In summary, we have ρ(c, c 2 ) (c c 2 ) 2 = ρ(g g 2 ) (kg g 2 ) 2. Emloying these relations, we have the following uer bound for the inner sum E(k) #(Λ g k XR (0) Ψ k ) Y ρ(g, g 2 ) (kg g 2 ) 2. (g,g 2 ) T : kg g 2 <Y α In order to be able to aly the level of distribution formula, we slit the summation into dyadic intervals. Given g and g 2 such that kg g 2 < Y α, there exist unique integers n and m such that 2 n g < 2 n, and 2 m g 2 < 2 m. Hence k2 n 2 m kg g 2 < Y α. We arrive at the estimate E(k) #(Λ g k XR (0) Ψ k ) Y ρ(g, g 2 ). n,m: k2 n+m <4Y α (g,g 2 ) T : g <2 n g 2 <2 m (kg g 2 ) 2 The inner sum is amenable to the level of distribution formula, and we have that E(k) is bounded above by a quantity of order: 2 n,m: k2 n+m <4Y α n+m (log 2 n+m+ ) ν + Y /2 k (2n+m ) /2 (log 2 n+m+ ) ν 2. To calculate this, we introduce the quantity Q := log 2 (Y α /k). Then E(k) 0 n<q 0 m<q n 0 n<q + Y /2 k 2 n (n + ) ν 0 n<q 2 n+m (n + m + ) ν + Y /2 0 m<q n 2 n/2 (n + ) ν 2 2 m (m + ) ν 0 m<q n k (2n+m ) /2 (n + m + ) ν 2 2 m/2 (m + ) ν 2. 29
30 Now if β > 0, θ, and N, then 2 tβ (t + ) θ β N θ 2 Nβ. 0 t<n Alying this result to our estimate for E(k), we arrive at: E(k) 2 Q 0 n<q (n + ) ν (Q n) ν + (2Q Y ) /2 2 Q Q 2ν+ + (2Q Y ) /2 Q 2ν2+. k k 0 n<q Recalling the definition of Q, we have the uer bound: E(k) Y α (log Y )ν k for some absolute constant ν. Finally, we sum E(k) over k: + Y (α+)/2 k 3/2 (log Y ) ν, (n + ) ν 2 (Q n) ν 2 E ε Y ε Y α (log Y ) ν + + Y ε Y (α+)/2 (log Y ) ν ε Y max(α+ε,α/2+/2+ε). If we choose ε = min(( α)/5, α /2), then E α Y/(log Y ) 3 and condition (C) is satisfied for any α < Conditions (D) and (E) For a A, one has that a = q (x)q 2 (x) with x Ψ. The set Ψ was chosen so that for all x Ψ, (q (x); D) = (q 2 (x); D) =, so (a; D) =, whence (a; P) =, satisfying condition (D). In the consideration of condition (E), we observe that for all a A, one has a X 4 Y 2. That is, there exists a constant C (deending only on the choice of forms q and q 2 ) such that a CY 2 for all a A. Define θ by C = Y θ. In order to satisfy condition (E), we need a Y µα, and it is sufficient to chose α < and µ such that µ (2 + θ)/α. A more careful analysis is required if we wish to make use of Diamond and Halberstam s exlicit result that I(2,, 2) > 5. By continuity of I, there exists η > 0 such that I(2, α, µ) > 5, rovided that α, µ 2 < η. Set µ = (2 + θ)/α. For α <, the above condition translates into α > (2 + θ)/(2 + η) and α > η. We can choose such a value of α rovided that θ < η. Now θ = log C/ log Y, so the condition will be satisfied for all sufficiently large Y. 30
31 3.2.5 Alication of Theorem 3. Having verified the conditions of Theorem 3., we find that for sufficiently large X, there exists a constant ν > 2 such that ( {P 5 : P 5 A} X 2 ω() ), <X 2/v and this is sufficient for the roof of Theorem.. Before we conclude, let us consider the condition in Theorem. that there exists z such that (q i (z); D) = for i =, 2. The condition is not always satisfied, as the following air of forms demonstrate: q (x, y) = 3x 2 + 2xy + y 2, q 2 (x, y) = 2x 2 4xy + 3y 2. Here, at least one of q (x, y) and q 2 (x, y) is divisible by 3 for every choice of x and y, and 3 divides D. On the other hand, we exect that the condition will be satisfied for most airs of forms and we exhibit the following infinite class of forms for which the condition holds: q (x, y) = x 2 + 2b xy + c y 2, q 2 (x, y) = x 2 + 2b 2 xy + c 2 y 2. Take z = (, 0), then q i (z) =, satisfying the condition as long as the resultant Res(q, q 2 ) is nonzero. 3.3 Conclusion Our investigations into airs of binary quadratic forms deended crucially on deriving an aroriate level of distribution formula and then alying the weighted sieve of Diamond and Halberstam. This technique is not limited to airs of forms and could be extended to the consideration of arbitrarily many binary quadratic forms. The level of distribution formula would give rise to arameters κ, α, and µ, and the main comutational roblem would be the calculation of the number r in Theorem 3.. Acknowledgements I d like to thank Roger HeathBrown for his unstinting guidance and suervision throughout this research. Thanks, too, to Glyn Harman and Tim Browning for their many helful suggestions. A final debt of gratitude is owed to Trevor Wooley, who originally osed this roblem. 3
32 References [] Duncan A. Buell, Binary quadratic forms, SringerVerlag, New York, 989, Classical theory and modern comutations. [2] David A. Cox, Primes of the form x 2 + ny 2, A WileyInterscience Publication, John Wiley & Sons Inc., New York, 989, Fermat, class field theory and comlex multilication. [3] Stehan Daniel, On the divisorsum roblem for binary forms, J. Reine Angew. Math. 507 (999), [4] Harold Davenort, Multilicative number theory, second ed., Graduate Texts in Mathematics, vol. 74, SringerVerlag, New York, 980, Revised by Hugh L. Montgomery. [5] H. Diamond and H. Halberstam, Some alications of sieves of dimension exceeding, Sieve methods, exonential sums, and their alications in number theory (Cardiff, 995), London Math. Soc. Lecture Note Ser., vol. 237, Cambridge Univ. Press, Cambridge, 997, [6] D. R. HeathBrown, Linear relations amongst sums of two squares, Number theory and algebraic geometry, London Math. Soc. Lecture Note Ser., vol. 303, Cambridge Univ. Press, Cambridge, 2003, [7] M. N. Huxley, A note on olynomial congruences, Recent Progress in Analytic Number Theory, Vol. (H. Halberstam and C. Hooley, eds.), Academic Press, London, 98, [8], Area, lattice oints, and exonential sums, London Mathematical Society Monograhs. New Series, vol. 3, The Clarendon Press Oxford University Press, Oxford, 996, Oxford Science Publications. [9] Henryk Iwaniec, Almostrimes reresented by quadratic olynomials, Invent. Math. 47 (978), no. 2, [0] E. C. Titchmarsh, The theory of the Riemann zetafunction, second ed., The Clarendon Press Oxford University Press, Oxford, 986, Edited and with a reface by D. R. HeathBrown. Mathematical Institute, St Giles, Oxford, 32
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