ON SIMULTANEOUS DIAGONAL INEQUALITIES, II

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1 ON SIMULTANEOUS DIAGONAL INEQUALITIES, II SCOTT T. PARSELL. Introduction In this paper we continue the investigation begun in []. Let λ,...,λ s µ,...,µ s be real numbers, define the forms F (x) =λ x λ s x 3 s, G(x) =µ x µ sx 2 s. Further, let τ be a positive real number. Our goal is to determine conditions under which the system of inequalities F (x) <τ, G(x) <τ (.) has a non-trivial integral solution. As has frequently been the case in work on systems of diophantine inequalities (see for example Brüdern Cook [6] Cook [7]), we were forced in [] to impose a condition requiring certain coefficient ratios to be algebraic. A recent paper of Bentkus Götze [4] introduced a method for avoiding such a restriction in the study of positive-definite quadratic forms, these ideas are in fact flexible enough to be applied to other problems. In particular, Freeman [0] was able to adapt the method to obtain an asymptotic lower bound for the number of solutions of a single diophantine inequality, thus finally providing the expected strengthening of a classical theorem of Davenport Heilbronn [9]. The purpose of the present note is to apply these new ideas to the system of inequalities (.). Write x =max( x,..., x s ), let N(P ) denote the number of integral solutions of the system (.) satisfying x P. We establish the following result. Theorem. Suppose that s 3, letλ,...,λ s µ,...,µ s be real numbers such that for some i j both of the ratios λ i /λ j µ i /µ j are irrational. Then one has N(P ) P s 5, provided that (i) the form F (x) has at least s 4 variables explicit, (ii) the form G(x) has at least s 5 variables explicit, (iii) the system of equations F (x) =G(x) =0has a non-singular real solution. In [], we actually provided a more quantitative conclusion, in which the parameter τ in (.) was replaced by an explicit function of x. Specifically, it was shown that when the coefficients of F G are algebraic then under the conditions of Theorem the system F (x) < x σ, G(x) < x σ 2 (.2) 99 Mathematics Subject Classification. D75 (D4, P55).

2 2 SCOTT T. PARSELL has at least on the order of P s 5 σ σ 2 integer solutions in the box x P, provided that σ + σ 2 < /2. Results of this nature do not appear to be obtainable in the current state of knowledge without the restriction to algebraic coefficients. The reason is that the permissible size of σ + σ 2 is determined by the amount of excess savings one generates on the minor arcs, but the Bentkus-Götze-Freeman method produces minor arc estimates of the form o(p s 5 ) without giving any further indication of the actual order of magnitude. Our proof of Theorem is modeled on the paper of Freeman [0]. As usual, the main challenge is to hle the minor arcs. We first seek to demonstrate that large Weyl sums yield good rational approximations to the coefficients of F G. This is essentially a theorem of R. Baker [] (see also [2] [3]), but we require a slightly sharper version, which we establish in 2. Then in 3 we are able to apply a two-dimensional version of Freeman s argument, combined with the methods of [], to complete the proof. It should be noted that much of the analysis underlying the results quoted from [] dates to the work of Wooley [3], [4] on simultaneous additive equations. The author is grateful to Eric Freeman for alerting him to the work of Bentkus Götze [4], for supplying a preprint of his own paper [0], for helpful discussions of these important new ideas. 2. Diophantine Approximation Via Large Weyl Sums As usual, we adopt the notation e(z) =e 2πiz. From a result of Baker [], we know that whenever the exponential sum F (α) = e(α 3 x 3 + α 2 x 2 ) x P is large, one obtains good simultaneous rational approximations to the coefficients α 2 α 3. Unfortunately, the bound for the denominator of these approximations fails by a factor of P ε to allow us to initiate a minor arc analysis along the lines of Freeman [0]. Therefore we provide the following slight refinement, in the spirit of [0], Lemma 2. Lemma 2.. Let ε be a positive real number, suppose that P is sufficiently large in terms of ε. Suppose that F (α) γ /8 P,whereP /64 γ. Then there is a positive integer q, integers a 2 a 3 satisfying (q, a 2,a 3 )=, absolute constants c 0,c 2, c 3, such that q c 0 γ 65, qα 2 a 2 c 2 γ 2 P 2+ε, qα 3 a 3 c 3 γ 9 P 3. Proof. We follow Baker s argument fairly closely, deviating only as necessary to save the factor of P ε in the bound for q. In view of the conclusions of the lemma, we may assume throughout that ε is sufficiently small. We have by a trivial extension of Freeman [0], Lemma 2, that there exists a positive integer r, an integer b, absolute constants C 0 C 3 satisfying (b, r) =, r < C 0 γ 64, rα 3 b <C 3 γ 8 P 3. (2.) Applying Weyl differencing, we find that F (α) 2 S(h), h <P

3 where ON SIMULTANEOUS DIAGONAL INEQUALITIES, II 3 S(h) =S(h; α) = x P h e(3hα 3 x 2 +(3h 2 α 3 +2hα 2 )x). (2.2) Trivially, we have S(h) P for all h, so S(h) ( 2 5 γ/4 P + ) P 2 γ/4 P 2 h 5 γ/4 P for P sufficiently large, thus S(h) F (α) 2 2 γ/4 P 2 2 γ/4 P 2. (2.3) 5 γ/4 P< h <P Now the number of divisors of r is O(r /256 ), so on using (2.) (2.3) we find that 2 γ/4 P 2 S(h) cγ /4 S(h) d r 5 γ/4 P< h <P (h,r)=d 5 γ/4 P< h <P (h,r)=d for some D dividing r some absolute constant c. It follows that S(h) (2c) γ /2 P 2. 5 γ/4 P< h <P (h,r)=d Moreover, on putting C =(8c), we see that the terms for which S(h) Cγ /2 P contribute at most 2Cγ /2 P 2 to this sum, so on writing B = {h : 5 γ/4 P< h <P, (h, r) =D, S(h) >Cγ /2 P }, we find that S(h) 2Cγ /2 P 2, h B it follows that card(b) 2Cγ /2 P. (2.4) Choose any h B, put b 3 =3hb. Thensince h P we have by (2.) that 3hα 3 r b 3 < 3C 3 γ 8 P 2 < P 64 for P sufficiently large. Furthermore, we have S(h) >Cγ /2 P>r /2 P ε/6, on choosing ε sufficiently small. Therefore, we may apply Baker s final coefficient lemma ([2], Lemma 4.6) to obtain an approximation to the coefficient of the linear term in (2.2). Writing d =(r, b 3 ), we can find a positive integer t 8 such that td 3hα 3 r b 3 C 2 γ P 2+ε (2.5) trd (3h 2 α 3 +2hα 2 ) C 2 γ P +ε. (2.6) We note for future reference that D d D d 3D.

4 4 SCOTT T. PARSELL Now write x = x(h) =thd θ =2rα 2. On writing z for the nearest integer to xθ, we find using (2.5) (2.6) that xθ z dd 2trd hα 2 3 trd (3h 2 α 3 +2hα 2 ) +3htd 3hα 3 r 6C 2 γ P +ε. Put X =8P ζ =6C 2 γ P +ε.ash runs through the set B, thevalueofd is fixed, there are only 8 possible values for t, so (2.4) shows that we generate R distinct, non-zero values of x, where R 8 card(b) 4 Cγ/2 P. (2.7) Then for P sufficiently large we have R>24ζX, so by Lemma 4 of Birch Davenport [5] (see also Baker [2], Lemma 5.2) it must be the case that the ratio z/x is constant as h runs through B. Therefore, we can find integers u v, independent of h, with(u, v) =,such that z/x = u/v for all values of x corresponding values of z. Further, we can ensure that v is positive. Since u v are coprime, we must have v x for all x. But 5 γ/4 PD x 8PD, (2.8) so it follows that R 6P (vd) hence by (2.7) we see that Now for all h Bwe have by (2.7), (2.8), (2.9) that vd 64C γ /2. (2.9) vθ u = v x xθ z 5vDγ /4 P ζ 920C 3 γ 7/4 P 2+ε. Finally, we set q =2vr, a 2 = u, a 3 =2vb. Then (2.) (2.9) give wherewehavesetc 0 = 28C 0 C. Furthermore, we have q 2vDr c 0 γ 64 2, (2.0) qα 2 a 2 = vθ u c 2 γ 7/4 P 2+ε (2.) qα 3 a 3 =2v rα 3 b < 2C 3 vγ 8 P 3 c 3 γ 8 2 P 3, (2.2) where c 2 = 920C 3 c 3 =48C 3 C. If (q, a 3,a 2 ), then we may divide out the common factor still retain the inequalities (2.0), (2.), (2.2) above. The lemma therefore follows on recalling that γ. We remark that Lemma 2. sharpens Baker s theorem only in cases where F (α) is nearly of order P, which is the situation that arises in our application. Baker s original theorem gives good results for sums as small as P 3/4+ε, this fact having been thoroughly exploited in our earlier paper [].

5 ON SIMULTANEOUS DIAGONAL INEQUALITIES, II 5 3. The Davenport-Heilbronn Method We now recall the analytic set-up introduced in []. We may assume (after rearranging variables) that the first m of the µ i are zero, that the last n of the λ i are zero, that the remaining h = s m n indices have both λ i µ i nonzero. Then when s 3 we have by conditions (i) (ii) of Theorem that 0 m 5, 0 n 4, h 4. (3.) Furthermore, we may suppose that λ I /λ J µ I /µ J are irrational, where I = m + h 2 J = m + h. We may also assume that τ =, since this case may then be applied to the forms τ F (x) τ G(x) to deduce the general result. When P R are positive numbers, let A(P, R) ={n [,P] Z : p n, p prime p R} denote the set of R-smooth numbers up to P. Write α =(α 3,α 2 ), define generating functions F i (α) =F i (α; P )= e(λ i α 3 x 3 + µ i α 2 x 2 ) f i (α) =f i (α; P, R) = x P x A(P,R) e(λ i α 3 x 3 + µ i α 2 x 2 ). It will also be convenient to write g i (α 3 )=f i (α 3, 0) H i (α 2 )=F i (0,α 2 ). Further, we set R = P η. From now on, whenever a statement involves ε R, it is intended to mean that the statement holds for all ε>0, provided that η is sufficiently small in terms of ε. Finally, we assume throughout that P is chosen to be sufficiently large. According to Davenport [8], there exists a real-valued even kernel function K of one real variable such that K(α) min(, α 2 ) (3.2) = 0, if t, ˆK(t) = e(αt)k(α)dα [0, ], if t, (3.3) =, if t. 3 for all real numbers t. Weset K(α) =K(α 3 )K(α 2 ). Now let N(P ) be the number of solutions of the system (.) satisfying x i A(P, R) (i =,...,m+ h 3) x i P (i = m + h 2,...,s).

6 6 SCOTT T. PARSELL By using (3.3), one finds that where F(α) = m+h 3 i= N(P ) f i (α), H(α) = m+h i=m+h 2 F(α)G(α)H(α)K(α) dα, (3.4) F i (α), G(α) = s i=m+h+ F i (α). Before describing the dissection of the plane into major, minor, trivial arcs, we need the following two lemmas, which are straightforward extensions of the ideas of Freeman [0], Lemmas 3 4. Lemma 3.. Suppose that S j T j are fixed real numbers satisfying 0 <S j T j (j =2, 3). Then whenever {i, j} = {2, 3}, one has lim P sup S j α j T j α i R F I (α; P )F J (α; P ) P 2 =0. Proof. Let us first suppose that i = 2j = 3. For notational convenience, we write (α 3,α 2 )=(α, β). If the result fails to hold, then we can find ε>0, a sequence of positive real numbers {P n } tending to, a sequence of ordered pairs {α n } = {(α n,β n )} with S 3 α n T 3 β n R (n Z + ), such that F I (α n ; P n )F J (α n ; P n ) εpn 2 for all positive integers n. On making a trivial estimate, it follows that for each n one has F i (α n ; P n ) εp n (i = I,J). Whenever n is large enough so that P n ε 52, we may apply Lemma 2. with γ = ε 8.Thus we obtain integers q in a in satisfying q in c 0 ε 520 λ i α n q in a in c 3 ε 72 Pn 3 (i = I,J). (3.5) It follows that for n sufficiently large one has a in c 0 λ i T 3 ε c 3 ε 72, hence there are only finitely many possible 4-tuples (a In,q In,a Jn,q Jn ). So there must be a 4-tuple (a I,q I,a J,q J ) that occurs for infinitely many of the α n. The compactness of [S 3,T 3 ] then ensures that among these α n we can find a subsequence {α nl } converging to a non-zero limit α 0.Wehave λ I α nl q I a I c 3 ε 72 Pn 3 l λ J α nl q J a J c 3 ε 72 Pn 3 l for each l, so on letting l we find that λ I /λ J = a I q J /(a J q I ), contradicting the assumption that λ I /λ J is irrational.

7 ON SIMULTANEOUS DIAGONAL INEQUALITIES, II 7 For the case where i =3j = 2, we repeat the same argument except that in (3.5) we use the second inequality of Lemma 2. instead of the third eventually contradict the irrationality of µ I /µ J. Lemma 3.2. There exist positive, real-valued functions S j (P ) T j (P ), depending only on λ I,λ J,µ I,µ J, such that lim j(p )=0 P lim j(p )= P (j =2, 3) whenever {i, j} = {2, 3} one has lim P sup S j (P ) α j T j (P ) α i R F I (α; P )F J (α; P ) P 2 =0. Proof. Fix j = 2 or 3. Then for every positive integer m, Lemma 3. tells us that there is a real number P m = P m,j such that F I (α; P )F J (α; P ) P 2 m whenever P P m m α j m, we may clearly assume that the sequence {P m } is non-decreasing. Now when P satisfies P m P<P m+, we define S j (P )= m T j(p )=m. It follows easily that F I (α; P )F J (α; P ) P 2 m when P P m S j (P ) α j T j (P ), this suffices to complete the proof. We are now ready to describe the dissection of the plane that will be used to evaluate the integral (3.4). Let T j (P ) be as in Lemma 3.2, define the trivial arcs by t = {α : α 3 >T 3 (P ) or α 2 >T 2 (P )}. Write λ =8max λ i µ =8max µ i, i s i s define M = {α : α 3 λ P 2 α 2 µ P } to be the major arc. Finally, the minor arcs are given by m = R 2 \ (t M). As in [] [3], we have for any set n R 2 that F(α)G(α) dα f i (α) h 3 g j (α 3 ) m H k (α 2 ) n dα (3.6) n n for some i, j, k satisfying m + i m + h, j m, m + h + k s. (3.7)

8 8 SCOTT T. PARSELL With the abbreviations f = f i (α), g = g j (α 3 ), H = H k (α 2 ), we find using (3.) that f h 3 g m H n P ( s 3 f 0 + f 6 H 4 + g 6 H 4 + f 4 g 6). (3.8) In order to employ this decomposition, we need several mean value estimates. In each one, we are required to obtain the full savings of P 5 with fewer than 3 variables present, as the minor-arc bounds for F I F J stemming from Lemma 3.2 do not provide quantifiable gains over the trivial estimates. Lemma 3.3. Suppose that i, j, k satisfy (3.7) that m + h 2 l m + h. Then for any t>8/3 any unit square U =[c, c +] [d, d +], one has (i) F l (α) t f i (α) 0 dα P t+5, (ii) (iii) (iv) U U U U F l (α) t f i (α) 6 H k (α 2 ) 4 dα P t+5, F l (α) t g j (α 3 ) 6 H k (α 2 ) 4 dα P t+5, F l (α) t f i (α) 4 g j (α 3 ) 6 dα P t+5. Proof. We dissect U into major minor arcs as follows. Let M(q, a, b) ={α U: λ l α 3 q a <P 9/4 µ l α 2 q b <P 5/4 }, write M = M(q, a, b). 0 a,b q<p 3/4 (q,a,b)= Then by Baker [2], Theorem 5., one has F l (α) P 3/4+ε whenever α U\M. Therefore, by part (i) of [], Lemma 5 (see also [4], Theorem 2), one has F l (α) t f i (α) 0 dα P (3/4+ε)t P 7/3+ε P t+5 U\M for ε sufficiently small, since t > 8/3. Similar minor arc bounds for the integrals in (ii) (iv) follow by using parts (ii) (iv) of [], Lemma 5. For the major arcs, we again illustrate the argument by focusing attention on the integral in part (i). By Hölder s inequality, one has /3 2/3 F l (α) t f i (α) 0 dα F l (α) 3t dα f i (α) 5 dα, M M the result now follows on making a change of variables using [], Lemma 8, together with part (v) of [], Lemma 5. Estimates for the major arc integrals in (ii) (iv) follow in an identical manner on using parts (vi) (viii) of [], Lemma 5. U

9 ON SIMULTANEOUS DIAGONAL INEQUALITIES, II 9 The trivial arcs are now quite easy to hle. Since H(α) F I (α) 3 + F J (α) 3 + F K (α) 3, (3.9) where K = m + h, we find from (3.2), (3.6), (3.8), Lemma 3.3 that F(α)G(α)H(α)K(α) dα (T 2 (P ) + T 3 (P ) )P s 5, t since T j (P ) we see that this is o(p s 5 ). Let us now tackle the minor arcs. We first subdivide m into two regions. Let S j (P ) P be as in Lemma 3.2, put m = {α m : α 3 S 3 (P ) or α 2 S 2 (P )} m 2 = m \ m. We know from Lemma 3.2 that sup F I (α)f J (α) = o(p 2 ). (3.0) æ m Now we need a similar result on the set m 2. The basic idea is that if α m 2 F I (α) is large, then λ I α 3 µ I α 2 have good rational approximations, yet both are already close to zero when P is large, since S j (P ) 0. We may therefore hope to get a contradiction by showing that α must then in fact lie in the major arc. Suppose that α m 2 that F I (α) γ /8 P,where γ =(max{s 2 (P ),S 3 (P )}) /66. Since S j (P ) P we have γ P /66, hence Lemma 2. applies. Thus we obtain integers q, a 2,a 3,with(q, a 2,a 3 )=, such that q c 0 γ 65, µ I α 2 q a 2 c 2 γ 2 P 2+ε, λ I α 3 q a 3 c 3 γ 9 P 3. It follows that a 3 c 3 γ 9 P 3 + λ I α 3 q γ 9 P 3 + γ 65 S 3 (P ) P 2 + S 3 (P ) /66, similarly a 2 c 2 γ 2 P 2+ε + µ I α 2 q γ 2 P 2+ε + γ 65 S 2 (P ) P + S 2 (P ) /66, whence a 2 = a 3 =0whenP is sufficiently large. Therefore we have α 3 γ 9 P 3 α 2 γ 2 P 2+ε. For sufficiently large P, this implies that α M hence gives a contradiction. We therefore conclude that sup F I (α) γ /8 P = o(p ). (3.) æ m 2 Now we are ready to complete the minor arc analysis. By (3.2) (3.9), we have for some l with m + h 2 l m + h some unit square U that F(α)G(α)H(α)K(α) dα sup F I (α)f J (α) /8 F l (α) /4 F(α)G(α) dα. æ m m U Therefore by (3.6), (3.8), (3.0), (3.), Lemma 3.3 we have F(α)G(α)H(α)K(α) dα sup F I (α)f J (α) /8 P s 2/4 = o(p s 5 ). æ m m

10 0 SCOTT T. PARSELL The treatment of the major arc is almost identical to that of [], so our discussion will be somewhat brief. As usual, we must prune back to a smaller set N on which we can obtain asymptotics for the sums f i (α). We let W =(logp ) /4 define N = {α : α 3 WP 3 α 2 WP 2 }. Then by using Hölder s inequality, together with Lemma 9.2 of Wooley [3] Lemma 3.3, we find that F(α)G(α)H(α)K(α) dα P s 5 W σ M\N for some σ>0. It may be worth mentioning that Freeman [0] is able to avoid pruning entirely in his work on a single inequality. The factor of P ε in our estimate for qα 2 in Lemma 2. is what prevents us from extending the m 2 analysis down to the boundary of N in the α 2 direction. When α N, we are able to approximate F i (α) f i (α) by the functions v i (α) = P 0 e(λ i α 3 γ 3 + µ i α 2 γ 2 ) dγ P ( ) log γ w i (α) = ρ e(λ i α 3 γ 3 + µ i α 2 γ 2 ) dγ R log R as in []. Here ρ(x) denotes Dickman s function (see for example Vaughan [2], chapter 2). Thus we are able to show that F(α)G(α)H(α)K(α) dα J(P ), where J(P )= N ( m+h 3 i= w i (α) )( s i=m+h 2 denotes the singular integral. By arguing as in [], we find that J(P ) P s ˆK(F (γ)p 3 ) ˆK(G(γ)P 2 ) dγ, B ) v i (α) K(α) dα where B =[R/P, ] m+h 3 [0, ] n+3. Now by condition (iii) of Theorem the argument of Lemma 6.2 of Wooley [3], we may assume that there is a non-singular solution η to the equations F = G =0suchthatη lies in the interior of B when P is sufficiently large. By the inverse function theorem, we are then able to find a set V R 2 containing the origin, with meas(v ), such that J(P ) P s ˆK(z j P 3 ) ˆK(z k P 2 ) dz. V It now follows from (3.3) that J(P ) P s 5, this completes the proof of Theorem.

11 ON SIMULTANEOUS DIAGONAL INEQUALITIES, II References [] R. C. Baker, Weyl sums diophantine approximation, J. London Math. Soc. (2) 25 (982), [2], Diophantine inequalities, Clarendon Press, Oxford, 986. [3], Correction to Weyl sums diophantine approximation, J. London Math. Soc. (2) 46 (992), [4] V. BentkusF. Götze, Lattice point problems distribution of values of quadratic forms, Ann. of Math. (2) 50 (999), [5]B.J.BirchH.Davenport,On a theorem of Davenport Heilbronn, ActaMath.00 (958), [6] J. Brüdern R. J. Cook, On pairs of cubic diophantine inequalities, Mathematika38 (99), [7] R. J. Cook, Simultaneous quadratic inequalities, ActaArith.25 (974), [8] H. Davenport, On indefinite quadratic forms in many variables, Mathematika3 (956), 8 0. [9] H. Davenport H. Heilbronn, On indefinite quadratic forms in five variables, J. London Math. Soc. 2 (946), [0] D. E. Freeman, Asymptotic lower bounds for diophantine inequalities, Mathematika (to appear). [] S. T. Parsell, On simultaneous diagonal inequalities, J. London Math. Soc. (2) 60 (999), [2] R. C. Vaughan, The Hardy-Littlewood method, second edition, Cambridge University Press, Cambridge, 997. [3] T. D. Wooley, On simultaneous additive equations II, J. reine angew. Math. 49 (99), [4], On simultaneous additive equations IV, Mathematika45 (998), Department of Mathematics, Texas A&M University, College Station, Texas address: parsell@alum.mit.edu