ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d 4) ≥ MANOJVERMA 4 1 Abstract. Fork≥2,wederiveanasymptoticformulaforthenumberofzerosoftheforms 0 Qki=1(x22i−1+x22i)+Qki=1(x22k+2i−1+x22k+2i)−x24kk+1 2 and n x1Qki=1(x22i+x22i+1)+x2k+2Qki=1(x22k+2i+1+x22k+2i+2)−x24kk++13 a inthebox1≤xi≤P usingthecirclemethod. J 9 Mathematics SubjectClassification2010: Primary11D45;Secondary11D85, 11P55. ] T N . h 1. Introduction t a m The problems of representation of positive integers and of non-trivial representation of [ zero by forms with integral coefficients have attracted attention of many mathematicians 1 since the beginning of the 20th century. Most of the work in this area has been carried v out using the circle method. The square-root barrier implies that the circle method is 6 unlikely to yield an asymptotic formula in case of a form of degree d with less than 2d+1 6 3 variables. In 1962 Birch, Davenport and Lewis [1] found an asymptotic formula for the 2 number representations of zero in a box x P by the cubic forms in seven variables of i . | | ≤ 1 the type 0 N (x ,x ,x )+N (x ,x ,x )+a x3 4 1 1 2 3 2 4 5 6 7 7 1 where N and N are norm forms of some cubic field extensions of Q and a is a nonzero 1 2 7 : integer. Incidentally, whereas an asymptotic formula for the number of representations v i of an integer as a sum of eight cubes has been established by Vaughan [4], an asymptotic X formula in the case of a sum of seven cubes seems to be out of reach as of now. However, r a a lower bound of the expected order of magnitude has been established by Vaughan [5]. A few more families of cubic forms in seven variables have been dealt with. However, for d 4 no example of an asymptotic formula for the number of representations of either ≥ zero or of large positive integers by a form of degree d in 2d+1 variables seems to exist in the literature. In this paper, for each d 4, we establish an asymptotic formula for the ≥ numberof zeros of aparticular form of degree din 2d+1variables ina suitableexpanding box using the circle method. It turns out that our proof goes through in cases d = 1,2,3 as well and retrieves previously known results in those cases. For d even, say d = 2k, k 1, we look at the number zeros of the form ≥ k k f(x) = f(x ,...,x )= (x2 +x2 )+ (x2 +x2 ) x2k 1 4k+1 2i−1 2i 2k+2i−1 2k+2i − 4k+1 i=1 i=1 Y Y while for d odd, say d= 2k+1, k 0, we look at the number of zeros of the form ≥ k k f(x) = f(x ,...,x )= x (x2 +x2 )+x (x2 +x2 ) x2k+1 1 4k+3 1 2i 2i+1 2k+2 2k+2i+1 2k+2i+2 − 4k+3 i=1 i=1 Y Y 1 2 MANOJVERMA in a box 1 x P. When convenient we shall abbreviate and write i ≤ ≤ f(x) = f(x ,...,x )= f (x ,...,x )+f (x ,...,x ) xd (1.1) 1 2d+1 d 1 d d d+1 2d − 2d+1 where k (x2 +x2 ) if d is even, d= 2k, k 1 fd(x1,...,xd) = x i=1k 2(ix−21 +x22i ) if d is odd, d = 2k+1,≥k 0 . (cid:26) Q1 i=1 2i 2i+1 ≥ We could define fd = fd(x1,.Q..,xd) recursively: f1(x1) = x1, f2(x1,x2) = x21 +x22, and f (x ,...,x ) = (x2 +x2) f (x ,...,x ) for d 2. d 1 d d−1 d · d−2 1 d−2 ≥ Taking x = x = x = x = ... = 1, x = x = x = x = d−2 2d−2 d−4 2d−4 d−3 2d−3 d−5 2d−5 ... = 0 and applying the four square theorem we see that this form represents every integer infinitely many times; our interest lies in establishing the asymptotic formula for the number of representations of zero with 1 x P. Only minor changes are required i ≤ ≤ to prove the expected asymptotic formula for the number of representations of a large positive integer N by this form (or by the form obtained by changing the sign in the front of xd2d+1 to +) with the variables restricted so that 1 ≤ xi ≤ ⌊Nd1⌋. − Note that if d 3 then the form f is singular (e.g., consider the point (1, 0, 0, ..., 0)) ≥ (soDeligne’s estimate for complete exponential sumsis notapplicable) andits h-invariant is 3. (The h-invariant of a form F, denoted h(F), is defined as the smallest integer h such that F can be written in the form F(x) = A (x)B (x)+...+A (x)B (x) 1 1 h h where A (x) and B (x) are forms of positive degrees having rational coefficients.) i i Inwhatfollows, ε can take any positive real valuein any statement in which itappears, e(α) = e2πiα, the symbols and O have their usualmeanings with the implicit constants ≪ depending at most on ε and d. If A is a set, #A denotes the cardinality of A. If n,m are positive integers, d(n) denotes the number of positive divisors of n and d (n) denotes m the number of (ordered) factorizations of n with m positive integral factors. For a real numberx, x and x denote thegreatest integer less than or equal to x andthe smallest ⌊ ⌋ ⌈ ⌉ integer greater than or equal to x respectively. Theorem 1. The number of zeros R(0;P) of the form f in (1.1) with 1 x P satisfies i ≤ ≤ R(0;P) = Pd+1SJ +O(Pd+1−1+5·12d−1+ε) where ∞ q a S = q−2d−1S(q,a) with S(q,a) = e f(z) , q q=1 a=1 zmodq (cid:18) (cid:19) X X X (a,q)=1 and ∞ J = e(γf(ξ))dξ dγ; Z−∞ Z[0,1]2d+1 ! S and J are positive. Remark. One can also consider the problem of finding an asymptotic formula for the number of zeros of the form f inside the box x P but then the contribution from i | | ≤ those “degenerate” zeros of f that lie inside the box x P and also in the set i | | ≤ U = x Z2d+1 :f (x ,...,x ) = f (x ,...,x )= x = 0 d 1 d d d+1 2d 2d+1 { ∈ } would be 22d−2P2d−2+O (P2d−3) in case d is odd and 22d−6d2P2d−4+O (P2d−5) in case d d d is even and thus the contribution from the zeros of f with at least one of the x equal to i zero would be Pd+2 if d 5 and cannot be determined precisely via the circle method. ≫ ≥ ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 3 On the other hand, the number of zeros of f with x P but none of the x equal to i i | | ≤ zero is 22d+1 times the number of zeros with 1 x P in case d is even, and is 22d−1 i ≤ ≤ times the sum of the number of zeros with 1 x P of the three forms i ≤ ≤ f (x ,...,x )+f (x ,...,x ) xd , d 1 d d d+1 2d − 2d+1 f (x ,...,x )+f (x ,...,x ) xd − d 1 d d d+1 2d − 2d+1 and f (x ,...,x ) f (x ,...,x ) xd ; d 1 d − d d+1 2d − 2d+1 theorem 1 applies to the latter two forms as well. 2. The Minor Arcs Let P be a large positive integer, I = 1,2,...,P , B = I2d+1 and { } F(α) = e(αf(x)). (2.1) x∈B X Then the number of zeros of the form f inside the box 1 x P is i ≤ ≤ 1 1 R(0;P) = F(α)dα = F (α)2F (α)dα (2.2) 1 2 Z0 Z0 where F (α) = e(αf (x ,...,x )) 1 d 1 d x1,.X..,xd∈I and F (α) = e( αxd). 2 − x∈I X Let δ be a sufficiently small positive real number to be specified later. For 1 q Pδ, 1 a q, (a,q) = 1, define the major arcs M(q,a) by ≤ ≤ ≤ ≤ a M(q,a) = α : α Pδ−d − q ≤ (cid:26) (cid:12) (cid:12) (cid:27) (cid:12) (cid:12) and let M be the union of the M(q,a). The(cid:12)set m =(cid:12) (Pδ−d,1+Pδ−d] M forms the minor (cid:12) (cid:12) \ arcs. Lemma 2.1. We have 1 F (α) 2dα Pd+ε. 1 | | ≪ Z0 Proof. We assume that d is even, d = 2k, k 1; the proof is similar in case d is odd. We ≥ have 1 F (α)2dα = a(n)2 1 | | Z0 0≤nX≤2kPd where a(n)= # (x ,...,x ) Id : f (x ,...,x )= n for n 0. 1 d d 1 d { ∈ } ≥ Clearly a(0) = 1 and a(n) = Pd. For n 1, in any solution of k (x2 + 0≤n≤2kPd ≥ i=1 2i−1 x2 ) = n we must have x2 +x2 = n (1 i k) for some factorization n = n ...n 2i P 2i−1 2i i ≤ ≤ Q 1 k of n with k (positive) factors. For a positive integer r, the number of solutions of the 4 MANOJVERMA equation s2+t2 = r in non-negative integers s,t is 8d(r) rε. Thus, for 1 n 2kPd ≤ ≪ ≤ ≤ we have a(n) nε...nε = d (n)nε nεnε P2dε. ≪ 1 k k ≪ ≪ n1.X..nk=n Hence 1 F (α)2dα = 1+ a(n)2 1+P2dε a(n) Pd+2dε. (cid:3) 1 | | ≪ ≪ Z0 1≤nX≤2kPd 1≤nX≤2kPd Lemma 2.2. We have d+1+ε− δ F(α)dα P 2d−1. m ≪ Z Proof. For α m, by Dirichlet’s theorem on Diophantine approximation, there exist a,q ∈ with (a,q) = 1,q Pd−δ and α a q−1Pδ−d < q−2. Since α m (Pδ−d,1 Pδ−d), ≤ − q ≤ ∈ ⊆ − we must have 1 a < q and(cid:12) q >(cid:12)Pδ (otherwise α would be in M(q,a)). By Weyl’s ≤ (cid:12) (cid:12) inequality (See [2] or [6]), (cid:12) (cid:12) 1+ε− δ F2(α) P 2d−1. ≪ This, with (2.2) and lemma 2.1 gives F(α)dα = F2(α)F (α)dα sup F (α) F (α)2dα (cid:12)Zm (cid:12) (cid:12)Zm 1 2 (cid:12) ≤ (cid:18)α∈m| 2 |(cid:19)(cid:18)Zm| 1 | (cid:19) (cid:12) (cid:12) (cid:12) (cid:12)1 (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) sup F2(α) (cid:12)(cid:12) F1(α)2dα Pd+1+ε−2dδ−1. (cid:3) ≤ (cid:18)α∈m| |(cid:19)(cid:18)Z0 | | (cid:19) ≪ 3. The Major Arcs Todealwiththemajorarcsweneedtoobtainanapproximationtothegeneratingfunction F(α), (2.1), in terms of the auxiliary functions a S(q,a) = e f(z) (3.1) q zmodq (cid:18) (cid:19) X and I(β) = e(βf(ξ))dξ. (3.2) Z[0,P]2d+1 Lemma 3.1. For α M(q,a), writing α = (a/q)+β, we have ∈ F(α) = q−2d−1S(q,a)I(β)+O(P2d+2δ). ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 5 Proof. We have a a F +β = e +β f(x) q q (cid:18) (cid:19) x∈B (cid:18)(cid:18) (cid:19) (cid:19) X a = e +β f(qy+z) q zmodq y:qy+z∈B (cid:18)(cid:18) (cid:19) (cid:19) X X a = e f(qy+z) e(βf(qy+z)) q zmodq y:qy+z∈B (cid:18) (cid:19) X X a = q−4k−1 e f(z) q2d+1e(βf(qy+z)). q zmodq (cid:18) (cid:19)y:qy+z∈B X X The inner sum is a Riemann sum of the function e(βf(ξ)) over a 2d + 1-dimensional cuboid which depends on z. We want to replace it by the integral over the 2d + 1- dimensional cube [0,P]2d+1 and estimate the error in doing so. Firstly, any first-order partial derivative of e(βf(ξ)) is of the form 2πiβg(ξ)e(βf(ξ)) where g(ξ) is a degree d 1 form (in 2d+1 real variables) whose coefficients are bounded and which itself is, − therefore, O(Pd−1) on the cube [0,P]2d+1. Therefore, e(βf(ξ)) does not vary by more that O(β qPd−1) on a cube of side q. There are q2d+1 terms in the outer sum and | | (P/q)2d+1 terms in each inner sum so the error in replacing the sum by the integral ≪ over the corresponding cuboid is O(β qPd−1q2d+1(P/q)2d+1) = O(β qP3d). Secondly, | | | | each side of this cuboid depends on z but is within length q of the sides of [0,P]2d+1. Since the integrand is bounded by 1 in absolute value, the error in replacing the region of integration by [0,P]2d+1 is O(qP2d). Since we have q Pδ and β Pδ−d, the total ≤ | | ≤ error is O(P2d+2δ)+O(P2d+δ) = O(P2d+2δ). Hence, a a F +β = q−2d−1 e f(z) I(β)+O(P2d+2δ) q q (cid:18) (cid:19) zmXodq (cid:18) (cid:19)(cid:16) (cid:17) which gives the lemma. (cid:3) Lemma 3.2. We have F(α)dα = Pd+1S(Pδ)J(Pδ)+O(Pd+5δ) M Z where q S(Q) = q−2d−1S(q,a) (3.3) q≤Q a=1 X X (a,q)=1 and J(µ) = e(γf(ξ))dξ dγ. (3.4) Z|γ|<µ Z[0,1]2d+1 ! Proof. Note that M(q,a) = [(a/q) Pδ−d,(a/q)+Pδ−d] has length 2Pδ−d. The change − of variable α= (a/q)+β and lemma 3.1 give Pδ−d F(α)dα = q−2d−1S(q,a) I(β)dβ +O(Pd+3δ). ZM(q,a) Z−Pδ−d 6 MANOJVERMA Summing over 1 q Pδ, 1 a q, (a,q) = 1, we get ≤ ≤ ≤ ≤ Pδ−d F(α)dα = S(Pδ) I(β)dβ +O(Pd+5δ). ZM Z−Pδ−d Applying the change of variable γ = Pdβ, the integral in the above equation becomes Pδ P−d I(P−dγ)dγ. Z−Pδ Since f is a homogeneous function of degree d, the change of variable ξ′ = P−1ξ gives (P−dγ) = e(γP−df(ξ))dξ = e(γf(P−1ξ))dξ Z[0,P]2d+1 Z[0,P]2d+1 = P2d+1 e(γf(ξ′))dξ′ Z[0,1]2d+1 which completes the proof of the lemma. (cid:3) 4. The Singular Integral Note that the inner integral appearingin the definition of J(µ), (3.4), is boundedtrivially by 1. Applying the change of variable ξ′ = γ1/2kξ we see that e(γf(ξ))dξ = γ−2dd+1 e(f(ξ′))dξ′. 1 Z[0,1]2d+1 Zγd[0,1]2d+1 We think of the last integral as an iterated integral and apply Lemma 7.3 of Vaughan [6] to each iteration to get the bound e(γf(ξ))dξ min(1,γ−2dd+1). (4.1) Z[0,1]2d+1 ≪ Lemma 4.1. For µ 1, ≥ J(µ) = J +O(µ−d+d1) where ∞ J = e(γf(ξ))dξ dγ >0. Z−∞ Z[0,1]2d+1 ! Proof. The convergence of J(µ) to J as µ at the rate asserted follows from (4.1). → ∞ To show that J > 0, we imitate [2, Chapter 4]. Writing α for (α ,...,α ) and α for 1 1 d 2 (α ,...,α ), d+1 2d λ J = lim e(γ(f (α )+f (α ) αd )dα dα dα dγ. λ→∞Z−λ Z[0,1]2d+1 d 1 d 2 − 2d+1 1 2 2d+1! Since 01e(−γαd2d+1)dα2d+1 = 01 d1αd1−1e(−γα)dα and −λλe(γµ)dγ = sin(2πλµ)/πµ for any real numbers λ and µ, R R R J = d1 λl→im∞Z[0,1]2d+1α1d−1(cid:18)sin(π2π(fλd((fαd1()α+1)f+d(fαd2()α−2)α−)α))(cid:19)dα1dα2dα. ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 7 Putting β = f (α )+f (α ) α so that α = f (α )+f (α ) β, d 1 d 2 − d 1 d 2 − 1 2⌊d/2⌋+1 sin(2πλβ) J = lim φ(β) dβ dλ→∞Z−1 πβ where φ(β) = d1 Z{β 0. (cid:3) 5. The Singular Series From the definition of S(Q), (3.3), q S(Q) = S(q) = q−2d−1 S (q,a)2S (q,a) (5.1) 1 2 q≤Q q≤Q a=1 X X X (a,q)=1 where q q a a S (q,a) = e f (x ,...,x ) , S (q,a) = e xd 1 d 1 d 2 q −q x1,.X..,xd=1 (cid:18) (cid:19) Xx=1 (cid:18) (cid:19) and q S(q) = q−2d−1 S (q,a)2S (q,a). 1 2 a=1 X (a,q)=1 S(q) is a multiplicative function of q. In fact, if (q ,q ) = (a,q q ) = 1 and we choose 1 2 1 2 a ,a so that a q +a q = 1 then for i= 1,2 1 2 2 1 1 2 S (q q ,a) = S (q q ,a q +a q ) = S (q ,a )S (q ,a ). (5.2) i 1 2 i 1 2 1 2 2 1 i 1 1 i 2 2 This follows from the fact that as r and r run through a complete (or reduced) set of 1 2 residues modulo q and q respectively, r q +r q runs through a complete (or reduced) 1 2 2 1 1 2 set of residues modulo q q . Thus it suffices to study S (q,a) when q is a prime power 1 2 i and (a,q) = 1. For a prime power pl and p ∤ a we have pl pl auy2 pl auz2 S (pl,a) = N (p,l,u) e e . (5.3) 1 d−2  pl pl  u=1 y=1 (cid:18) (cid:19)z=1 (cid:18) (cid:19) X X X   where N (p,l,u) = # (x ,...,x ): 1 x ,...,x pl,f (x ,...,x ) u(mod pl) . d−2 1 d−2 1 d−2 d−2 1 d−2 { ≤ ≤ ≡ } For a fixed u with pj u, writing u = pju and dividing the sum over y (or z) into pj 1 k subsums by grouping pl−j consecutive terms together we get pl auy2 pl auz2 pj pl−j au (pl−jv +v )2 pl−j au v2 e = e = e 1 1 2 = pj e 1 2 . pl pl pl−j pl−j Xy=1 (cid:18) (cid:19) Xz=1 (cid:18) (cid:19) vX1=1vX2=1 (cid:18) (cid:19) vX2=1 (cid:18) (cid:19) 8 MANOJVERMA The sum over v is a quadratic Gauss sum modulo pl−j and its value is well known (e.g., 2 see [3, Chapter 9]). We find that p2l if j = l pl auy2 pl auz2  pl+j if p ≡ 1(mod 4) e e = ( 1)l+jpl+j if p 3(mod 4) = G(p,l,j) , Xy=1 (cid:18) pl (cid:19)Xz=1 (cid:18) pl (cid:19)  0− if p =≡ 2, j =l−1 2 √ 1 2l+j if p = 2, j l 2  · − · ≤ −  say, depends on p,l and j but not on a or u1. Thus, from (5.3), l−1 S (pl,a) = p2lM (p,l,l)+ M∗ (p,l,j)G(p,l,j) (5.4) 1 d−2 d−2 j=0 X where, for d 3, l 1, 0 j l, ≥ ≥ ≤ ≤ M (p,l,j) := # (x ,...,x ): 1 x ,...,x pl, pj f (x ,...,x ) . d−2 1 d−2 1 d−2 d−2 1 d−2 { ≤ ≤ | } and, for d 3, l 1, 0 j l 1, ≥ ≥ ≤ ≤ − M∗ (p,l,j) := # (x ,...,x ) :1 x ,...,x pl, pj f (x ,...,x ) d−2 { 1 d−2 ≤ 1 d−2 ≤ k d−2 1 d−2 } = M (p,l,j) M (p,l,j +1). d−2 d−2 − The equation (5.4) also holds for d= 1 and d= 2 if we define M (p,l,j) =0 for 0 j l, M∗ (p,l,j) = 0 for 0 j l 1, −1 ≤ ≤ −1 ≤ ≤ − 1 if j = 0 1 if j = 0 f = 1, M (p,l,j) = and M∗(p,l,j) = . 0 0 0 if 1 j l 0 0 if 1 j l 1 (cid:26) ≤ ≤ (cid:26) ≤ ≤ − It is clear that M (p,l,j) = pl−j for 0 j l and M∗(p,l,j) = pl−j(1 p−1) for 0 j l 1. 1 ≤ ≤ 1 − ≤ ≤ − The treatment of the singular series is trivial in cases d = 1,2,3 now; we dispose them off. When d = 1 we have S (pl,a) = 0; S (q,a) = S (q,a) =S(q) = 1 if q = 1, 0 if q > 1; S= 1. (5.5) 1 1 2 When d = 2 we have S1(pl,a) = G(p,l,0); S1(q,a) 2q; S2(q,a) 2q; S(q) 4√2q−23. (5.6) | | ≤ | | ≤ | | ≤ When d = 3 we have p S1(pl,a) < (l+1)p2l; S1(q,a) < q2d(q); S2(q,a) q2/3; S(q) q−43+ε. (5.7) | | | | | ≪ | | ≪ To deal with the cases d 4, we first calculate M (p,l,j) = # (w,x) : 1 w,x 2 ≥ { ≤ ≤ pl, pj w2 + x2 for 1 j l. We do so by dividing the set (w,x) : 1 w,x | } ≤ ≤ { ≤ ≤ pl, pj w2+x2 according to the highest power of p dividing w. Thus, for 1 j l, | } ≤ ≤ M2(p,l,j) = # (w,x) :1 w,x pl, p⌊j+21⌋ w, pj w2+x2 { ≤ ≤ | | } ⌊j−1⌋ 2 + # (w,x) : 1 w,x pl, pm w, pj w2+x2 { ≤ ≤ k | } m=0 X ⌊j−1⌋ 2 = p2l−2⌊j+21⌋+ # (w′,x′): 1 w′,x′ pl−m, p ∤w′, pj−2m w′2+x′2 . { ≤ ≤ | } m=0 X Since 1 has two square roots modulo pj−2m if p 1 (mod 4), one square root modulo − ≡ pj−2m if pj−2m = 2, and no square root in other cases, the mth term in the sum equals ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 9 pl−m(1 1) 2 p(l−m)−(j−2m) = 2p2l−j(1 1) if p 1 (mod 4), 22l−j−1 if pj−2m = 2 and − p · · − p ≡ zero otherwise. Thus, for 1 j l, ≤ ≤ p2l−2⌊j+21⌋+2 j+1 p2l−j(1 1) if p 1(mod 4) ⌊ 2 ⌋ − p ≡ M2(p,l,j) =  p2l−2⌊j+21⌋ if p 3(mod 4) .  22l−j if p =≡ 2   Hence, for 0 j l 1, ≤ ≤ − (j +1)p2l−j(1 1)2 if p 1(mod 4) − p ≡ p2l−j(1 1 ) if p 3(mod 4) and j is even M∗(p,l,j) =  − p2 ≡ . 2  0 if p 3(mod 4) and j is odd  22l−j−1 if p =≡ 2    Thus for 0 j l 1 we have M∗(2,l,j) = 22l−1−j, M∗(p,l,j) < p2l−j if p 3(mod 4), ≤ ≤ − 2 2 ≡ and M∗(p,l,j) < (j +1)p2l−j for all p. Also, M (p,l,l) pl if p = 2 or p 3(mod 4), 2 2 ≤ ≡ and M (p,l,l) < (l+1)pl for all p. To estimate S (q,a) we consider the cases of even d 2 1 and odd d separately. Now suppose that d is even and d 4, say d = 2k and k 2. Then, for 0 j l 1, ≥ ≥ ≤ ≤ − k−1 M∗ (p,l,j) = M∗(p,l,j ) 2k−2 2 i 0≤Xji≤l−1 Yi=1 j1+...+jk−1=j Thus k−1 M∗ (2,l,j) = 22l−1−ji 2k−2 0≤Xji≤l−1 Yi=1 j1+...+jk−1=j = 2(k−1)(2l−1)−j 0≤Xji≤l−1 j1+...+jk−1=j lk−12(k−1)(2l−1)−j lk−122kl−2l−1−j. ≤ ≤ If p 3(mod 4), then ≡ k−1 M∗ (p,l,j) < p2l−ji 2k−2 0≤Xji≤l−1 Yi=1 j1+...+jk−1=j = p2(k−1)l−j 0≤Xji≤l−1 j1+...+jk−1=j lk−1p2(k−1)l−j = lk−1p2kl−2l−j. ≤ 10 MANOJVERMA If p 1(mod 4), then ≡ k−1 M∗ (p,l,j) < (j +1)p2l−ji 2k−2 i 0≤Xji≤l−1 Yi=1 j1+...+jk−1=j k−1 = p2(k−1)l−j (j +1) i 0≤Xji≤l−1 Yi=1 j1+...+jk−1=j < lk−1p2(k−1)l−j 1 0≤jXi≤l−1 j1+...+jk−1=j < lk−1lk−1p2(k−1)l−j = l2k−2p2kl−2l−j. Dividing the (2k 2)-tuples (x ,...,x ) with pl f (x ,...,x ) into the ones 1 2k−2 2k−2 1 2k−2 − | with one of the factors x2 +x2 divisible by pl and the ones with none of the factors 2i−1 2i x2 +x2 divisible by pl we see that for any prime p 2i−1 2i k−1 k 1 M (p,l,l) − M (p,l,l) (p2l)k−2+ M∗(p,l,j ) 2k−2 ≤ 1 2 · 2 i (cid:18) (cid:19) 0≤Xji≤l−1 Yi=1 j1+...+jk−1≥l < (k 1)(l+1)pl p2kl−4l+l2k−2p2kl−2l−l − · < (l+1)2k−2p2kl−3l. (The last inequality follows since (k 1)(l+1)+l2k−2 (k 1)(2l)+l2k−2 = 2k−2 l+ − ≤ − 1 · l2k−2 < (l+1)2k−2.) Thus from (5.4), (cid:0) (cid:1) S (pl,a) < (l+1)(l+1)2k−2p2kl−l = (pl)2k−1d(pl)2k−1. 1 | | Thus from (5.2) for even d 4 and (a,q) = 1, ≥ S (q,a) q2k−1d(q)2k−1 q2k−1+ε = qd−1+ε. (5.8) 1 | | ≤ ≪ Now supposethat d is odd and d 5, say d= 2k+1 and k 2. Note that there is one ≥ ≥ value of x modulo pl that is divisible by pl while for 0 j l 1 there are pl−j(1 p−1) 1 ≤ ≤ − − values of x modulo pl that are exactly divisible by pj. Hence, for 0 j l 1, 1 ≤ ≤ − k−1 M∗ (p,l,j) = pl−j0(1 p−1) M∗(p,l,j ). 2k−1 − 2 i 0≤Xji≤l−1 Yi=1 j0+...+jk−1=j Thus k−1 M∗ (2,l,j) = 2l−j0(1 2−1) 22l−1−ji 2k−1 − 0≤Xji≤l−1 Yi=1 j0+...+jk−1=j 1 = 2l+(k−1)(2l−1)−j 2 · 0≤Xji≤l−1 j0+...+jk−1=j lk2l+2kl−2l−(k−1)−j−1 lk22kl−l−j−2. ≤ ≤

On a form of degree $d$ in $2d+1$ variables ($d\geq 4$)

Similar On a form of degree $d$ in $2d+1$ variables ($d\geq 4$)