THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS

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1 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS R. C. BAKER Abstract. Let F(x 1,..., x n be a nonsingular inefinite quaratic form, n = 3 or 4. Results are obtaine on the number of solutions of F(x 1,..., x n = 0 with x 1,..., x n square-free, in a large box of sie P. It is convenient to count solutions with weights. Let ( x R(F, w = µ 2 (x w P F(x=0 where w is infinitely ifferentiable with comact suort an vanishes if any x i = 0, while µ 2 (x = µ 2 ( x 1... µ 2 ( x n. It is assume that F is robust in the sense that et M 1... et M n 0, where M i is the matrix obtaine by eleting row i an column i from the matrix M of F. In the case n = 3, there is the further hyothesis that et M 1, et M 2, et M 3 are not squares. It is shown that R(F, w is asymtotic to e n σ (F, wρ (FP n 2 log P, where e n = 1 for n = 4, e n = 1 2 for n = 3. Here σ (F, w an ρ (F are resectively the singular integral an the singular series associate to the roblem. The metho is aate from the aroach of Heath-Brown to the corresoning roblem with x 1,..., x n unrestricte integer variables. Let F(x = F(x 1,..., x n = n i, j=1 1. Introuction a i j x i x j (a i j = a ji Z be a nonsingular inefinite quaratic form, n 3. Let M = [a i j ], D = et(m. We are concerne here with the asymtotics of the square-free solutions x Z n, of (1.1 F(x = 0. As in [1], let For x Z n, let π y = y 1 y n (y R n. 0 if π x = 0 µ(x = µ( x 1... µ( x n if π x 0. A square-free solution of (1.1 is a solution having µ(x 0. Solutions of (1.1 will be weighte, as in [1], by a function w ( x P, where the ositive arameter P tens to infinity. We assume throughout that (i w is infinitely ifferentiable with comact suort; (ii w(x = 0 whenever π x = 0, 2000 Mathematics Subject Classification. Primary 11P55; Seconary 11E20. 1

2 2 R. C. BAKER (iii w(x 0, an w(x > 0 for some real solution x of (1.1. Our object of stuy is R(F, w = F(x=0 ( x µ 2 (xw. P An asymtotic formula for R(F, w was obtaine in [1] in the cases (a n 5, (b n = 4; D not a square. The metho use was an elaboration of that of Heath-Brown [4], whose objective was to obtain an asymtotic formula for ( x N(F, w = w. P F(w=0 Besies the cases (a, (b, Heath-Brown also successfully treate N(F, w in the more ifficult cases (c n = 4; D a square, ( n = 3. In the resent aer, I treat R(F, w for the cases (c, (. Some restrictions are imose on F. Let M j be the matrix obtaine by eleting row j an column j of M. We say that F is robust if (1.2 et(m 1... et(m n 0. Our results will aly to robust forms, with a further restriction when n = 3. In orer to state the asymtotic formulae, we efine the singular integral by 1 σ (F, w = lim w(xx, ɛ 0 + 2ɛ F(x ɛ where... x enotes integration over R n with resect to Lebesgue measure. Uner the conitions (i (iii, σ (F, w is ositive ([4], Theorem 3. The singular series for our roblem is ( ρ (F = 1 1 ρ. Here ρ is given by ρ = lim ν ν(n 1 #{x (mo ν : F(x 0 (mo ν, 2 x 1,..., 2 x n }. Thus ρ is the -aic ensity of solutions of F = 0 square-free with resect to. Theorem 1. Let n = 4, let D be a square an suose that F is robust. Then R(F, w = σ (F, wρ (FP 2 log P + O(P 2 log P(log log P 1+ɛ. As usual, ɛ is an arbitrary ositive number, suose sufficiently small. Constants imlie by O an may een on F, w an ɛ. Any other eenence will be shown exlicitly.

3 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 3 Theorem 2. Let n = 3 an suose that F is robust. Suose further that none of et M 1, et M 2, et M 3 is a square. Then R(F, w = 1 2 σ (F, wρ (FP log P + O(P log P(log log P 1/2. The following roositions give information about ρ (F. Proosition 1. Let F be nonsingular (if n = 4 an robust (if n = 3. (a if ρ > 0 for every rime, then ρ (F > 0. (b if the congruence F(x 0 (mo (2D 5 has a solution with 2 x 1,..., 2 x n whenever 2D, then ρ (F > 0. Proosition 2. If n = 3 an F is not robust, then ρ (F = 0. As an examle for Proosition 2, it is a simle exercise to show that P #{x : µ(x 0, P x j < 2P, F 0 (x = 0} P for the ternary form F 0 (x = 2x 1 x 2 2x3 2. The conclusion of Theorem 2 clearly extens to F 0! In fact, I conjecture that for a non-robust ternary quaratic form F an a given w, there is an asymtotic formula R(F, w c(f, wp with c(f, w > 0, recisely when w > 0 at some oint of a certain set E = E(F of zeros of F. In the examle, E = {(t, t, ± t : t 0}. Before outlining the roofs of Theorems 1 an 2, we recall some notations from [1] an [4]. We write, for c Z n, q S q,f (c = S q (c = e q (af(b + c b. a=1 As usual, the asterisk inicates (a, q = 1, while c b = c 1 b + c n b n, b (mo q e(θ = e 2πiθ, e q (m = e ( m. q The symbols an t are reserve for oints in Z n with ositive square-free coorinates. Let F (x = F( 2 1 x 1,..., 2 n x n an similarly for w (x. We write S q (, c = S q,f (c. It is convenient to write m as an abbreviation for Further, let 1 m,..., n m. y = max( y 1,..., y n.

4 4 R. C. BAKER Let h(x, y (x > 0, y R be the smooth function that occurs in Theorem 1 an 2 of [4]. We recall that h(x, y is nonzero only for x max(1, 2 y. It is shown in [4, Theorem 2] that N(F, w = c P P 2 q n S q (ci q (c, c Z n where (1.3 c P = O N (P N for every N > 0, an ( x I q,f,w (c = I q (c = w R n P Clearly I q,f,w (c is nonzero only for q P. As note in [1], (1.4 I q,f,w (c = 1 π 2 Thus (1.5 N(F, w = c P π 2 P2 Let c Z n ( q h P, F(x e P 2 q ( c xx. ( c1 I q,..., c n. 2 2 S q (, c q n z = z(p = 1 log log P, 7 Q(z =. <z I q c 1,..., c n. 1 2 n 2 For x Z n, π x 0, let 1 if 2 x j for < z an j = 1,..., n f z (x = 0 otherwise. It is easy to verify that f z (x µ 2 (x f z (x 1 z 2 x 1 1. z 2 x n Multilying by w ( x P an summing over x Z n with F(x = 0, ( x ( x (1.6 f z (xw R(F, w f z (xw P P Here F(x=0 (1.7 S j (X = F(x=0 (S 1 (z + + S n (z. ( x w. P X, 2 x j F(x=0 We note that f z (x = µ(, 2 1 x 1 1 Q(z 2 n x n n Q(z

5 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 5 so that (1.8 F(x=0 ( x f z (xw = P = Q(z F(x=0 µ(n(f, w. ( x w P 2 1 x 1,..., 2 n x n Q(z We can exress S j (X somewhat similarly. Take for examle j = 1 an write ( = (, 1,..., 1, F = F (, w = w (. Then (1.9 S 1 (X = N(F, w. X Our lan is to aat [4] so as to evaluate N(F, w via (1.5, making the error exlicit in, an then aly this to the last exression in (1.8 an to N(F, w. The contribution to S 1 (z from P ɛ will receive a more elementary treatment, similar to [1, Proosition 1]. In conclusion, I oint out a refinement of a theorem in [1] ue to Blomer [2]. Let R(m be the number of reresentations of m as a sum of 3 squarefree integers. If the square-free kernel of m is at least m δ, for a ositive constant δ, an m 1, 3 or 6 (mo 8, then Blomer obtains µ( (1.10 R(m = c S(mm 1/2 + O(m (1 γ/2, γ = γ(δ > 0. Here c is the singular integral an S(m the singular series, m ɛ S(m m ɛ. In [1], (1.10 is obtaine only for square-free m. 2. Some exonential integrals, exonential sums an Dirichlet series From now on we assume that n = 3 or 4, an the eterminant of F is a square for n = 4. It suffices to rove Theorems 1 an 2 for weight functions w with the following roerty: there exists a ositive number l = l(f, w such that, whenever (x 0, y su(w, we have F x (x, y 1 ( x x 0 l an F has exactly one zero (x, y with x x 0 l. We shall assume that w has this roerty. The euction of the general case of Theorems 1 an 2 is carrie out by a simle roceure given on age 179 of [4]. As note on age 180 of [4], (2.1 I q (v = P n I r (v (r = P 1 q, where (2.2 Ir (v = w(xh(r, F(xe r ( v xx. R n For v = 0, we have (2.3 I r (0 = σ (F, w + O N (r N for any N > 0, rovie that r 1 [4, Lemma 13]. Consequently (2.4 I q (0 = P n {σ (F, w + O N ((q/p N }

6 6 R. C. BAKER for q P. By combining the conclusions of [4, Lemmas 14, 15, 16, 18, 19, 22], we arrive at the following bouns: (2.5 I r (v 1, (2.6 (2.7 (2.8 (2.9 (2.10 (2.11 (2.12 r I r (v r 1, I q (v P n, q I q(c q P n, I r (v N r 1 v N (N 1 I q (v N P n+1 q 1 v N (N 1, I r (v (r 2 v ɛ (r 1 v 1 n/2, I q (v P n ( P 2 v q 2 ɛ ( 1 n/2 P v, q (2.13 Lemma 1. For any K > 1, (2.14 (2.15 (2.16 (2.17 q q I q(v P n ( P 2 v q ɛ ( 1 n/2 P v. q r 1 I r (vr K v K ( v > 1, ( 2 r 1 Ir (vr log v ( v 1, q 1 I q (vq M P n v K ( v > 1, ( 2 q 1 I q (vq P n log v ( v 1. Proof. In view of (2.1, it suffices to rove (2.14 an (2.15. Suose first that v > 1. We use (2.11 for the range r v N/2 an (2.9 for the remaining range. Thus 0 v N/2 r 1 Ir (vr v ɛ+1 n/2 r n/2 1 2ɛ r + v N v N/2 r 2 r 0 v ɛ+1 n/2 N/2 (n/2 2ɛ + v N/2 K v K

7 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 7 for a suitable choice of N = N(K, ɛ. Now suose that v 1. We use (2.11 for the range r v, (2.5 for the range v < r v 1, an (2.9 with N = 1 for the remaining range. Thus Thus 0 v r 1 Ir (vr v ɛ+1 n/2 r n/2 1 2ɛ r 0 v 1 + r 1 r + v 1 v v 1 ɛ + 2 log v 1 r 2 r ( log v We now turn to estimates for S q (, c. Let M be the matrix M = [ 2 i 2 j a i j]. (2.18 et M = π 4 et(m, et M 1 Writing M 1 = 1 et(m [b i j], so that b i j Z, we note that (2.19 M 1 1 b i j = et(m. 2 i 2 j = (et(m 1. π 4 ( 2. v We write M 1 (x for the quaratic form, with rational coefficients, whose matrix is M 1 Let = 2 et M. When π, we may think of M 1 (x as being efine moulo. We recall that, for any nonsingular form F, (2.20 S q (, c q 1+n/2 ( 2 1, q... (2 n, q [1, Lemma 9]. We nee a slight generalization of (2.20. Let c(a be a vector in Z n for every a = 1,..., q, (a, q = 1. Let q S = e q (af (b + c(a b. Then a=1 a=1 b (mo q S q 1+n/2 ( 2 1, q... (2 n, q. To see this, Cauchy s inequality yiels q S 2 φ(q e q (a(f (u F (v + c(a (u v. Substitute u = v + w, so that u,v (mo q e q (a(f (u F (v + c(a (u v = e q (af(w + c(a w e q (av F(w..

8 8 R. C. BAKER The summation over v will now rouce a contribution of zero unless q ivies F (w = 2M w. We have S 2 q n φ(q 2 1. w (mo q 2M w 0 (mo q We may now comlete the roof with the argument use for [1, Lemma 9]. Since (2.21 S uv (, c = S u (, vcs v (, ūc where uū 1 (mo v, v v 1 (mo u [4, Lemma 23], we can o most of our work for rime owers q. For n = 4, M 1 (c 0, we have (2.22 S q (, c π 2 X7/2+ɛ ( c + 1 ɛ q X [1, Lemma 10]. To get results that lay a comarable role when n = 4, M 1 (c = 0 or n = 3, we use the Dirichlet series ζ(s,, c = q s S q (c (σ > 2 + n/2 { = u=0 } us S u(, c [4,. 194]. Bouns for those Euler factors for which π will require extra work comare to the analysis on ages of [4]. If we write (2.23 τ (c, σ = uσ S u(, c, π we see that the analysis in question gives (i for n = 3, M 1 (c 0, (2.24 ζ(s,, c = L(s 2, χ,c ν(s,, c where ν(s,, c = u=0 { (1 χ,c ( 2 s an χ,c is a character satisfying (2.25 χ,c ( = et(m M 1 (c. u=0 } us S u(c, We note that χ,c (if not trivial is a character to moulus 4 π 4 M 1 (c. Moreover, (2.26 ν(s,, c c ɛ τ (c, σ (σ ɛ. (ii for n = 3, M 1 (c = 0, (2.27 ζ(s,, c = ζ(2s 5ν(s,, c,

9 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 9 with Moreover, { ν(s,, c = (1 5 2s (2.28 ν(s,, c τ (c, σ u=0 } us S u(, c. (σ ɛ. (iii for n = 4, M 1 (c = 0, we fin that ζ(s,, c = L(s 3, χ ν(s,, c, where { ν(s,, c = (1 χ ( 3 s u=0 } us S u(, c, with a character χ satisfying ( et M χ ( =. Since et M is a square, we take the trivial character, an write (2.29 ζ(s,, c = ζ(s 3ν(s,, c. Moreover, (2.30 ν(s,, c τ (c, σ (σ 72 + ɛ. For any, we write t j = t j ( for the rouct of those rimes iviing exactly j of 1,..., n. Eviently, π = t 1 t tn n. We also write π 5ɛ (2.31 A( = t2 2 (t 3t 4 4 (n = 4 π 5ɛ t5/2 2 t3 4 (n = 3. Let us write α 3 = 17/6, α 4 = 7/2. Lemma 2. (i Let F be nonsingular. Then τ (c, σ π 2+ɛ (σ α n + ɛ. (ii Let σ n ɛ. Suose that F is nonsingular (if n = 4 an robust (if n = 3. Then τ (c, σ A(. Proof. (i For n = 3, (2.20 yiels S u(, c σu (1/3+ɛu ( 2 1, 2 ( 2 2, 2 ( 2 3, 2, S u(, c σu ( 2 1, 2 ( 2 2, 2 ( 2 3, 2, τ (c, σ π ɛ (1 2, 2 (2 2, 2 (3 2, 2 π = π 2+ɛ.

10 10 R. C. BAKER The argument is similar for n = 4. (ii For n = 4, (2.20 yiels (2.32 If ivies t 1, then If ivies t 2, then If ivies t 3 t 4, then S u(, c σu (1 ɛ ( 2 1,... (2 4, S u(, c σu 4+4ɛ + + (2 2ɛ ( 2 1, 2 ( 2 4, 2. S u(, c σu 2ɛ. S u(, c σu 2+2ɛ. 8 (1 ɛu 4+4ɛ. Here we use the trivial boun (u 4 an (2.20 (u 5. Lemma 2(ii follows for n = 4. Now let n = 3. Suose that t 1 ; let us say 1. Then for u 4, an a fixe value of x 1, let us write G(x 2, x 3 = u 5 3 a jk 2 j 2 k x jx k, j,k=2 h = h(a = (aa x c 2, aa x c 3. We have 3 af (x c ag(x 2, x 3 + aa 1k k x 1x k + x c u S u(, c 2 u k=2 x 1 c ag(x 2, x 3 + (x 2, x 3 h (mo u, x 1 =1 u a=1 y (mo u e(ag(y + y h 2 (by Cauchy s inequality 2u ( 2u 2 by the generalization of (2.20 note above, with n relace by 2 an F relace by F(0, x 2, x 3. Hence, alying (2.20 irectly for u 5, S u(, c σu 4ɛ + 4ɛ. u 5 2 u( 1 2 ɛ

11 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 11 Now let t 2. Then (2.33 S u(, c σu 2+2ɛ (u 2 4 u(1/2 ɛ (u 3. Here we use the trivial boun (u 2 an (2.20 (u 3. Hence S u(, c σu 5/2+3ɛ. Similarly, if t 3, S u(, c σu 4+4ɛ (u 4 6 u( 1 ɛ 2 (u 5, S u(, c σu 4+4ɛ. We now comlete the roof as above. The next lemma is useful for singular series calculations. Lemma 3. Let F be nonsingular, Λ (F = nu S u(, 0. Then Λ (F 2 π >1 3/2 If n = 3 an F is not robust, then Λ (F 1. (n = 4, F nonsingular (n = 3, F robust. Proof. Suose first that n = 4. The roof of Lemma 2 (ii shows that, for, 1 (π = nu S u(, 0 2 (π = 2 4 (π 3. Hence π 2 nu S u(, 0 2, an we obtain the esire boun since has O(1 values. The argument for n = 3 is similar in the case when F is robust. However, if F is not robust, we have the weaker boun (2.34 nu S u(, 0 (π =. For the left-han sie of (2.34 is from ( /2 ( 2 1, (2 2, (2 3, + 1 ( 2 1, 2 ( 2 2, 2 ( 2 3, 2

12 12 R. C. BAKER 3. Sums of S q (, c an S q (, 0q n. Let e n = 1 if n = 4 an e n = 1/2 if n = 3. We assume throughout Sections 3 an 4 that F is robust (n = 3 an nonsingular (n = 4. Define e n if M 1 (c = 0 (3.1 η(, c = 1 if n = 3 an (et M M 1 (c is a nonzero square 0 otherwise. Define σ(, c = ν(n,, c. We observe that whenever η(, c 0, σ(, c = σ (, c, where σ (, c = (1 1 nu S u(, c. Lemma 4. For X > 1, S q (, c = η(, cσ(, c Xn n + O(Xα n+2ɛ π 3+ɛ ( c 1/2. q X Proof. The case n = 4, M 1 (c 0 follows from (2.22, an we exclue this case below. We recall the version of Perron s formula given in [1, Lemma 13]. Let b, c be ositive constants an λ a real constant, λ + c > b. For K > 0 an comlex numbers a l (l 1 with a l Kl b, write a l h(s = l s (σ > b; then (3.2 l x l=1 u=0 a l l λ = 1 c+it ( h(s + λ xs Kx c 2πi c it s s + O T whenever x > 1, T > 1, x 1/2 Z. For n = 4, let a l = S l (, c, b = 3, λ = 0, x = [X] + 1/2, T = x 10. Accoring to (2.20, we may take K π 2. Recalling (2.29, q X S q (, c = 1 2πi = 1 2πi 5+iT 5 it 5+iT 5 it ζ(s,, c x2 s + O(π2 ζ(s 3ν(s,, c xs s s + O(π2. We move the line of integration back to σ = ɛ. On the line segments [7/2 + ɛ, 5] ± i T, ζ(s 3 T 1/4, ν(s,, cx s s π 2+ɛ T 1/2

13 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 13 from (2.30 an Lemma 2 (i. Thus these segments contribute O(π 2+ɛ. Since U ( 1 L(σ + it, χ 2 t k 1/2 U 2 < σ < 1 0 for a Dirichlet L-function to moulus k, we have T ( 1 ζ 2 + ɛ + it ν(s,, c T t t π2+ɛ log T. Hence the segment [7/2 + ɛ it, 7/2 + ɛ + it] contributes O(X 7/2+2ɛ π 2+ɛ. Writing Res for the resiue of the integran at s = 4, with Res = 0 if there is no ole, S q (, c = Res +O(π 2+ɛ X 7/2+2ɛ. q X Similarly, for n = 3, S q (, c = 1 2πi where q X 5+iT 5 it ζ(s,, c xs s s + O(π2 = 1 5+iT E(sν(s,, c xs 2πi 5 it s s + O(π2, L(s 2, χ if M 1 (c 0 E(s = ζ(2s 5 if M 1 (c = 0 an χ = χ,c satisfies (2.25. We take χ to be the trivial character if et(m M 1 (c is a nonzero square. Since χ is a character to moulus k = O(π 4 c 2, a simle hybri boun [3, Lemma 1] yiels E(s = O((kT 1/4 = O ( ( c 1/2 π T 1/4 for σ 11/4, t T. We move the line of integration back to σ = 17/6 + ɛ. A slight variant of the receing argument gives S q (, c = Res +O ( X 17/6+2ɛ ( c 1/2 π 3+ɛ. q X It now suffices to show that the resiue at n is In the case n = 4, the resiue is as require. η(, cσ(, c xn n. ν(4,, c x4 4 For n = 3, there is no ole unless either M 1 that is, et(m M 1 (c = 0 or M 1 (c 0 an χ,c is trivial, x3 (c is a nonzero square. The resiue is σ(, c 3 or 1 x3 2 σ(, c 3 in the Laurent exansion of the zeta factor is 1 eening on whether the coefficient of s 3 1 or 1 2, an the lemma follows.

14 14 R. C. BAKER Lemma 5. For X > 1, q n S q (, 0 = e n σ(, 0 log X q X q X + O(A( + π 2+ɛ X α n n+2ɛ. Proof. For n = 4, we aly (3.2 with a l, b, x, T, K as in the receing roof, but now λ = 4, c = 1. This leas to q 4 S q (, 0 = 1 1+iT ζ(s + 1ν(s + 4,, 0 xs 2πi s s + O(π2. 1 it We move the line of integration back to σ = ɛ. The integrals along segments are O(π 2+ɛ X 1/2+2ɛ by a variant of the above argument. There is a ouble ole at 0; the Laurent series of the integran is 1 s 2 ( as + (ν(4,, 0 + ν (4,, 0s + ( (log xs +, where a is an absolute constant. The resiue is ν(4,, 0(log x + a + ν (4,, 0 ( = σ(, 0 log X + O max τ (0, σ + 1. σ 4 ɛ To get the last estimate, we write ν (4,, 0 as a contour integral on s 4 = ɛ using Cauchy s formula for a erivative, an aly (2.30. We now comlete the roof using Lemma 2 (ii. For n = 3, a similar argument gives q X q 3 S q (, 0 = 1 2πi 1+iT 1 it ζ(2s + 1ν(s + 3,, 0 xs s s + O(π2. We move the line of integration back to σ = ɛ, estimating the integrals along line segments as O(π 2+ɛ X 1/6+ɛ. This time the Laurent series at 0 is 1 2s ( 2as + (ν(3,, ν (3,, 0s + ( (log xs + with resiue 1 2 ν(3,, 0(log x + 2a ν (3,, 0, an we comlete the roof as before. We fix for the resent, with an write Lemma 6. We have c Z n c >P ɛ 4. Evaluation of N(F, w. c = P ɛ, c 1 2 1,..., c n. c 2 n q n S q (, ci q (c Pn.

15 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 15 Proof. We note first that for A 1, R > 1, N 2, (4.1 (ca 1 N = A N c>ar A N Taking A = 2 1, R = Pɛ, we have k=0 2 k AR<c 2 k+1 AR c N 2 (N 1k A N+1 R N+1 AR N+1. k=0 c 1 > 2 1 Pɛ (c 1 2 P 2(n 1ɛ 1 N ( c max 1 2 2,..., cn n 2 c c 1 > 2 1 Pɛ (c N+n 1 P 2(n 1ɛ 2 1 P (N nɛ P (N 3nɛ. Here we allow for a ossible renumbering of the variables. If N = N(ɛ is chosen suitably, we get the lemma by combining this estimate with (2.10 an (2.20, on recalling that the summation over q is restricte to q P. Lemma 7. Let c P ɛ. Then (4.2 q n S q (, ci q (c = η(, cσ(, c Proof. Let For R 1 2, (4.3 R<q 2R q n S q (, ci q (c = T(q = S l (, c, l q B = π 3+ɛ ( c 1/2. 2R R 0 q n I q (c T(q q n I q (c q + O(P α n+20ɛ. = q n I q (c 2R 2R T(q R R q (q n I q (c T(qq { } η(, cσ(, cq = q n I q (c n + O(Bq αn+2ɛ 2R n R 2R { } η(σ(, cq R q (q n I q (c n + O(Bq αn+2ɛ q n from Lemma 4. Now for R < q 2R, q n I q (c P n/2+1+2ɛ R n/2 1, q (q n I q (c P n/2+1+2ɛ R n/2 2

16 16 R. C. BAKER from (2.12, (2.13. Hence the O-terms in the last exression in (4.3 contribute O(BP n/2+1+2ɛ R n/2 1+αn+2ɛ. We conclue that (4.4 q n S q (, ci q (c = R<q 2R η(, cσ(, c 2R R q 1 I q (c q + O(BP n/2+1+2ɛ R n/2 1+α n+2ɛ. The lemma follows because q = O(P for the nonzero terms of the series in (4.2. Lemma 8. We have q n S q (, ci q (c Pn. c > ɛ Proof. By Lemma 6, we can restrict the sum to ɛ < c P ɛ. Let K > 1. Combining Lemma 7 with (2.16, these c contribute K P n c K σ(, c + P αn+24ɛ c > ɛ K P n c K+ɛ π 2+ɛ c > ɛ + P n by (2.26, (2.28, (2.30 an Lemma 2 (i. The last exression is (arguing as in the roof of Lemma 6 K P n+2nɛ π 2+ɛ (c K+n 1+ɛ + P n. c 1 >1 2 ɛ The lemma now follows from an alication of (4.1 with N = K n 1 ɛ, A = 2 1, R = ɛ ; K is suitably chosen eening on ɛ. Lemma 9. Let Then 0 < c ɛ. q n S q (, ci q (c P n A(η(, c + P αn+20ɛ. Proof. In view of Lemma 7 it suffices to show that σ(, c 0 q n I q (c q P n A(. The integral is P n log(2 by (2.16, (2.17 an the simle observation that c 2. The require estimate for σ(, c is rovie by (2.26, (2.28, (2.30 an Lemma 2 (ii (with ɛ/2 in lace of ɛ. It remains to treat the series q n S q (, 0I q (0. Lemma 10. We have q n S q (, 0I q (, 0 = e n σ(, 0σ (F, wp n log P + O(P n A(.

17 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 17 Proof. To begin with, (4.5 q n S q (, 0I q (0 q P 1 ɛ = (from (2.14 an (2.20 q P 1 ɛ q n S q (, 0P n σ (F, w + O N (π 2 Pn+(1 ɛ/2 P ɛn = e n σ(, 0σ (F, wp n log P 1 ɛ + O(P n A( by Lemma 5 together with an aroriate choice of N. For the range q > P 1 ɛ, we use (4.4. Cruely, (4.6 q n S q (, 0I q (c q>p 1 ɛ = e n σ(, 0 q 1 I q (0q + O(π 3+ɛ P n/2+1+2ɛ. P 1 ɛ Combining (4.5, (4.6, an substituting I q (0 = P n I r (0, where r = q/p, we obtain (4.7 Here q n S q (, 0I q (0 = e n σ(, 0σ (F, wp n log P L(λ = σ (F, w log λ + + e n σ(, 0L(P ɛ P n + O(P n A(. λ r 1 I r (0r. It is shown by Heath-Brown [4,. 203] that L(λ tens to a limit L(0 as λ tens to 0, an more recisely (4.8 L(λ = L(0 + O N (λ N. Recalling (2.28, (2.30 an Lemma 2 (ii, we see that (4.7 an (4.8 together yiel the lemma. Lemma 11. We have N(F, w = e n π 2 σ(, 0σ (F, wp n 2 log P + O(P n 2 π 2 A(#{c : c ɛ, η(, c 0}. Proof. Combining Lemmas 8, 9 an 10, an noting that 0 is counte in {c : c ɛ, η(, c 0}, q n S q (, ci q (, c c Z n = e n σ(, 0σ (F, wp n log P + O(P n A(#{c : c ɛ, η(, c 0}. The lemma now follows easily on combining this with (1.5 an (1.3.

18 18 R. C. BAKER 5. Comletion of the roof of Theorems 1 an 2. Lemma 12. Suose that F is nonsingular (n = 4 an robust (n = 3. In the notation of (1.7, we have S 1 (P ɛ P n. Proof. In view of (1.9, it suffices to show that (5.1 N(F, w P n+ɛ/2 2. This is a consequence of [1, Proosition 1] for n = 4. The roof of that roosition can be aate slightly to give (5.1 for n = 3. By following the argument on [1, ], we see that it suffices to show for 1 h P that the equation has O(P 1+ɛ/2 h 1 solutions with cx 2 z2 A 2z 2 2 = 0 (x, z 1, z 2 P, x 1 0, x 1 0 (mo h. Here c, A 2 are nonzero integers, since the quaratic form cx1 2 + z2 A 2z 2 2 is obtaine from F by a nonsingular linear change of variables. There are O(Ph 1 choices for x 1. For each of these, there are O(P ɛ/2 ossible (z 1, z 2 [1, Lemma 1]. This comletes the roof of the lemma. Lemma 13. Let F be nonsingular. Let B(q = t Q(z q µ( π 2 Then (i B(q is a multilicative function. µ(t (ii S π 2 q (t, 0 = B(q (1 2 n. t (iii For all rimes, <z q S q (, 0. (1 2 n nu B( u = (1 2 n ρ. u=1 Proof. (i This is a secial case of [1, Lemma 17]. (ii This is a variant of [1, Lemma 16]. The sum over t is unrestricte in [1]. (iii This is obtaine by letting N ten to infinity in the exression where N (1 2 n nu B( u = (1 2 n (n 1N M N, u=1 M N = #{x (mo N : F(x 0 (mo N, 2 x 1,..., 2 x n }, which is (5.9 of [1]. Convergence is a consequence of Lemma 3. Lemma 14. Let F be nonsingular (n = 4 an robust (n = 3. We have µ( (5.2 σ(, 0 = ρ (F + O((log log P e n. π 2 Q(z

19 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 19 Proof. Let ( V(w = 1 1. <w The left-han sie of (5.2 is lim w h(w, where h(w = Q(z = V(w µ( π 2 <w q <w ( 1 1 u=0 q n (after a simle maniulation. By Lemma 13 (ii, h(w = V(w Q(z µ( π 2 q n B(q (1 2 = V(w (1 2 1 <z 1 n 1 <z 1 q q <w S u(, 0 nu 1 n S q (, 0 C(q. Here C(q = q n B(q (1 2 1 <z 1 q 1 n is multilicative by Lemma 13 (i, an so h(w = V(w (1 2 1 n + C( + C( <w(1 2 + <z = (1 2 1 <z 1 n <w ( 1 1 ( a (z u=1 B( u nu. Here (1 2 n if < z a (z = 1 if z. Letting w ten to infinity, the left-han sie of (5.2 is (5.3 <z ( 1 1 ρ z ( 1 1 ( u=1 B( u nu

20 20 R. C. BAKER by Lemma 13 (iii. This is clearly close to ρ (F for large z. More recisely, ( 1 1 ( ρ = 1 1 ( (1 2 n + nu S u(0 = by Lemma 3. Now for 2D, ( 1 1 ( + π >1 µ( π 2 ( 1 1 ( u=1 u=1 nu S u(0 = as shown by Heath-Brown on. 195 of [4]; one takes 1 δ = 6 (n = (n = 4 in his argument. We obtain (5.4 u=1 S u(, 0 nu ( 1 1 ρ = O( (1+en. nu S u(0 + O( (1+c n O( 2 (n = 4 O( 3/2 (n = 3 Essentially the same argument shows that ( ( ( B( u nu = O( (1+en. u=1 It is now an easy matter to euce from (5.4 an (5.5 that the exression in (5.3 is ( 1 1 ρ + O(z e n as require. Proof of Proosition 1. Part (a is a straightforwar consequence of (5.4. For art (b, we may reeat verbatim the roof that ρ > 0 for all in [1, ]. Proof of Proosition 2. We nee only show that ( 1 1 ρ = 1 k ( 1 + O 3/2 where k is the number of j, 1 j 3, for which et M j = 0. Arguing as in the receing roof, this reuces to showing that µ( S u(, 0 (5.6 = k ( 1 π 2 3u + O. 3/2 π >1

21 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 21 Using unchange the art of the roof of Lemma 3 with π 2, we fin that these terms contribute O( 3/2 to the left-han sie of (5.6. A familiar argument also gives, for π =, S (, 0 1/2 ( 3u 1, ( 2, ( 3, u 2 + 3/2 ( 1, 2 ( 2, 2 ( 3, 2 1/2, so that terms with π =, u 2 also contribute O( 3/2. Write (1 = (, 1, 1, (2 = (1,, 1, (3 = (1, 1, It remains to show that S 2( ( j, 0 + O( 1/2 if et M j = 0 (5.7 = 6 O(1 if et M j 0. The case et M j 0 of (5.7 is essentially the same as the case n = 3, t 1 of the roof of Lemma 2 (ii. Now suose et M j = 0. Since M j has rank at least 1, its rank is 1. Taking j = 1 for simlicity of writing, [ ] [ ] x2 x2 x 3 M j = r x 3 s (bx 2 + cx 3 2 with integers r, s, b, c, rs 0, (b, c 0. For s, s s 1 (mo 2, Hence F( 2 x 1, x 2, x 3 r s(bx 2 + cx 3 2 (mo 2. S 2( ( j, 0 = 2 2 a=1 2 = 4 a=1 2 2 x 2 =1 x 3 =1 2 y=1 e (ay 2 e 2(ar s(bx 2 + cx 3 2 if rs(gc(b, c, since bx 2 + cx 3 takes each value (mo 2 exactly 2 times. The last exression is evaluate in [4, Lemma 27] as 4 2 ( 1 = 7 6, an the roof of the roosition is comlete. Lemma 15. Let F be nonsingular (n = 4 or robust (n = 3. Then (5.8 µ(n(f, w = e n σ (F, wρ (FP n 2 log P Q(z + O(P n 2 log P(log log P e n. Proof. From Lemma 11, the left-han sie of (5.8 is e n σ (F, wp n 2 µ(σ(, 0 (5.9 log P π 2 Q(z + O (P n 2 π 2 A(#{c : c ɛ }. The O-term is P n 2 Q(z Q(z π 2+4/3+2+ɛ 5ɛ

22 22 R. C. BAKER since A( = O(π 4/3+5ɛ. Moreover, for given k, #{ : Q(z}Q(z4k The O-term in (5.9 is thus Q(z π k Q(z 4k+ɛ e z(4k+2ɛ. P n 2 e 6z = P n 2 (log P 6/7. The lemma now follows on alying Lemma 14 to the first sum over in (5.9. Lemma 16. Uner the hyothesis of Theorem 1 or Theorem 2, we have z <P ɛ N(F, w = O(P n 2 log P(log log P 1+8ɛ. Proof. By Lemma 11, (5.10 Here N is the number of c in the box for which either N(F, w = e n 2 σ(, 0σ (F, wp n 2 log P + O(P n 2 2+5ɛ N. B : c 1 2+ɛ, c j ɛ (2 j n (5.11 et(m ( M 1 ( (c = 0 or n = 3 an (5.12 et(m ( M 1 ( (c = q2, for a nonzero integer q. Recalling (2.18, (2.19, we fin that with et(m ( M 1 ( (c = b 11c 2 22 n b 1 j c 1 c j + 4 j=2 b 11 = et(m 1 0. n b i j c i c j, We see at once that (5.11 hols for only O( 3ɛ oints c in B, since c 2,..., c n etermine c 1 to within two choices. If (5.12 hols, then (5.13 where 0 = b 11 et M ( M 1 ( (c + b 11q 2 n 2 = (b 11 c 2 b 1 j c j 4 l + b 11 q 2, j=2 ( n 2 l = b 1 j c j j=2 n b i j c i c j. i, j=2 i, j=2

23 THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS 23 If l = 0, then b 11 is a nonzero square from (5.13, in contraiction to the hyothesis of Theorem 2. We conclue that the ai of [1, Lemma 1] that for given c 2, c 3, (5.12 etermines c 1 to within O( ɛ ossibilities. Thus in all cases, N = O( 3ɛ. We use this estimate together with (2.28, (2.30 an Lemma 2 (ii to euce from (5.10 that N(F, w = O(P n 2 (log P 2+8ɛ. The lemma now follows. We are now reay to comlete the roofs of Theorems 1 an 2. From (1.6, (1.8, R(F, w µ(n(f, w n max S j (z n N(F, w Q(z after a ossible renumbering of the variables. Thus R(F, w = µ(n(f, w + O(P n 2 (log P(log log P 1+8ɛ Q(z (from Lemmas 12 an 16 = e n σ (F, wρ (FP n 2 log P + O(P n 2 (log P(log log P g n from Lemma 15. Here 1 8ɛ (n = 3 g n = 1/2 (n = 4. Since ɛ is arbitrary, this comletes the roof. References [1] R. C. Baker, The values of a quaratic form at square-free oints, Acta Arith. 124 (2006, [2] V. Blomer, Ternary quaratic forms, an sums of three squares with restricte variables, Anatomy of Integers, 1 17, Amer. Math. Soc., [3] D. R. Heath-Brown, Hybri bouns for Dirichlet L- functions, Invent. Math. 47 (1978, [4] D. R. Heath-Brown, A new form of the circle metho, an its alication to quaratic forms, J. Reine Angew. Math. 481 (1996, Deartment of Mathematics Brigham Young University Provo, UT 84602, USA baker@math.byu.eu j z