THE ZEROS OF A QUADRATIC FORM AT SQUARE-FREE POINTS

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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 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.

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 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

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 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 2 0 0 0 0 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

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 ( 1 + 1 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 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 194 5 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 176 + ɛ. (ii for n = 3, M 1 (c = 0, (2.27 ζ(s,, c = ζ(2s 5ν(s,, c,

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 176 + 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 t2 2... 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 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 12 2 1 2 2 x c 2, aa 13 2 1 2 3 x c 3. We have 3 af (x c ag(x 2, x 3 + aa 1k 1 2 2 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

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 (2.20. 1/2 ( 2 1, (2 2, (2 3, + 1 ( 2 1, 2 ( 2 2, 2 ( 2 3, 2

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 = 7 2 + 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

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 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 = 1 2 + 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 = 1 6 + 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,, 0 + 2 ν (3,, 0s + ( (log xs + with resiue 1 2 ν(3,, 0(log x + 2a + 1 2 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.

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 1 1 2 1 c 1 > 2 1 Pɛ (c 1 2 1 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 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 1 1 2 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(.

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 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,. 107 8], 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

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 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 = 3 1 2 (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 ( (5.5 1 1 ( 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,. 130 131]. 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

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 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

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, 101 137. [2] V. Blomer, Ternary quaratic forms, an sums of three squares with restricte variables, Anatomy of Integers, 1 17, Amer. Math. Soc., 2008. [3] D. R. Heath-Brown, Hybri bouns for Dirichlet L- functions, Invent. Math. 47 (1978, 149 170. [4] D. R. Heath-Brown, A new form of the circle metho, an its alication to quaratic forms, J. Reine Angew. Math. 481 (1996, 147 206. Deartment of Mathematics Brigham Young University Provo, UT 84602, USA Email: baker@math.byu.eu j z

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