Waring problem with the Ramanujan $\tau$-function

arXiv:math/0607169v1 [math.NT] 6 Jul 2006 Waring problem with the Ramanujan τ -function M. Z. Garaev, V. C. Garcia and S. V. Konyagin Abstract Let τ (n) be the Ramanujan τ -function. We prove that for any integer N the diophantine equation 74000 X τ (ni ) = N i=1 has a solution in positive integers n1 , n2 , . . . , n74000 satisfying the condition max ni ≪ |N |2/11 + 1. 1≤i≤74000 We also consider similar questions in the residue ring modulo a large prime p. 2000 Mathematics Subject Classification: 11B13, 11F35 1 Introduction The Ramanujan function τ (n) is defined by the expansion X ∞ Y n 24 (1 − X ) n=1 = ∞ X τ (n)X n . n=1 It possesses many remarkable properties of arithmetical nature. It is known that: • τ (n) is an integer-valued multiplicative function, that is, τ (nm) = τ (n)τ (m) if gcd(n, m) = 1; 1 • For any integer α ≥ 0 and prime q, τ (q α+2 ) = τ (q α+1 )τ (q) − q 11 τ (q α ). In particular, τ (q 2 ) = τ 2 (q) − q 11 . P 11 • τ (n) ≡ d (mod 691). d|n • τ (n) ≡ P d11 (mod 28 ) for n odd. d|n • |τ (q)| ≤ 2q 11/2 for any prime q and |τ (n)| ≪ n11/2+ε for any positive integer n and any ε > 0 (here and throughout the paper the constants implicit in the Vinogradov symbols “≪” and “≫” may depend only on ε); this has been proved by Deligne. For these and other properties of τ (n), see for example, [4], [7], [9], [12]. In particular, one can derive from [12] that any residue class modulo a prime number p can be expressed as τ (n) (mod p) for some positive integer n. Based on the deep sum-product estimate of Bourgain, Katz and Tao [3], and Vinogradov’s double exponential sum estimate, Shparlinski [13] established that the values τ (n), n ≤ p4 , form a finite additive basis modulo p, i.e., there exists an absolute integer constant s ≥ 1 such that any residue class modulo p is representable in the form τ (n1 ) + . . . + τ (ns ) (mod p) with some positive integers n1 , . . . , ns ≤ p4 . In the present paper, introducing a new approach, we prove a result which in a particular case reduces the number 4 on the exponent of p to the best possible 2/11. Theorem 1. The set of values of τ (n) forms a finite additive basis for the set of integers. Moreover, for any integer N the diophantine equation 74000 X τ (ni ) = N i=1 has a solution in positive integers n1 , n2 , . . . , n74000 satisfying the condition max ni ≪ |N|2/11 + 1. 1≤i≤74000 2 We remark that the quantity 74000 comes from 36 × 2050 + 200. Here the number 2050 (as well as the number 200) can be reduced if one uses the recent developments on the Waring-Goldbach problem, see [1, 10] and therein references. This however still gives a big number of summands. In this connection we prove the following results, where we assume that p is a large prime number. Theorem 2. For any integer λ the congruence 16 X i=1 τ (ni ) − 16 X i=1 τ (mi ) ≡ λ (mod p) holds for some positive integers n1 , m1 , . . . , n16 , m16 with max {ni , mi } ≪ p2 log4 p 1≤i≤16 and gcd(ni mi , 23!) = 1. Using Theorem 2, we show that any residue class modulo p is representable in the form 96 X τ (ni ) (mod p) i=1 2 4 with max ni ≪ p log p. In particular, for some positive constant C the set 1≤i≤96 {τ (n) (mod p) : n ≤ Cp2 log4 p} forms a finite additive basis for the residue ring Zp of order at most 96, see [8] for the definition. Theorem 3. For any integer λ and any ε > 0, the congruence 16 X i=1 τ (ni ) ≡ λ (mod p) is solvable in positive integers n1 , . . . , n16 with max ni ≪ p3+ε . 1≤i≤16 3 In particular, for any sufficiently large prime p the set {τ (n) (mod p) : n ≤ p3+ε } is a basis of Zp of order at most 16. Throughout the text the letters q, q1 , q2 , . . . , are used to denote prime numbers. For a given set A, |A| denotes its cardinality. Acknowledgement. The third author was supported by the INTAS grant 03-51-5070. 2 Preliminary statements First, we recall the following consequence of the classical result of Hua LooKeng [6]. Lemma 4. Let s0 be a fixed integer ≥ 2049 and let J denote the number of solutions of the Waring-Goldbach equation s0 X qi11 = N i=1 in primes q1 , . . . , qs0 with qi > 23 for all 1 ≤ i ≤ s0 . There exist positive constants c1 = c1 (s0 ) and c2 = c2 (s0 ) such that for all sufficiently large integer N, N ≡ s0 (mod 2), the following bounds hold: c1 N s0 /11−1 N s0 /11−1 ≤ J ≤ c . 2 (log N)s0 (log N)s0 Next, we require the following result of Glibichuk [5]. Lemma 5. If X , Y ∈ Zp with |X ||Y| > 2p, then ( 8 ) X xi yi : xi ∈ X , yi ∈ Y = Zp . i=1 4 We will also need some special computational results concerning the values of τ (n). Due to the aforementioned result of Deligne |τ (n)| ≪ n11/2+ε , the following N 6 numbers 6 X τ (ai ), i=1 1 ≤ a1 , . . . , a6 ≤ N are all of the size O(N 11/2+ε ). Thus, in average, each number is representable by the sum of six values of τ (n) many times. It is natural to expect that zero can also be expressed as a sum of six values of τ (n). With this in mind we search six positive integers a1 , . . . , a6 satisfying 6 X τ (ai ) = 0. i=1 There are many formulas which connect τ (n) with the function X σs (n) = ds . d|n It is known, for example, that n−1 65 691 691 X τ (n) = σ5 (k)σ5 (n − k). σ11 (n) + σ5 (n) − 756 756 3 k=1 Another formula looks like 4 τ (n) = n σ0 (n) − 24 n−1 X k=1 (35k 4 − 52k 3 n + 18k 2 n2 )σ0 (k)σ0 (n − k), see [11]. The formulas of the above type are useful for numerical computations of τ (n). In particular, one can extract that τ (12) = −370944, τ (27) = −73279080, τ (55) = 2582175960, τ (69) = 4698104544, τ (90) = 13173496560, τ (105) = −20380127040. 5 Now we have τ (12) + τ (27) + τ (55) + τ (69) + τ (90) + τ (105) = 0. (1) Therefore, assuming that Theorem 2 is proved, we multiply the set equality 16 nX i=1 τ (ni ) − 16 X i=1 o τ (mi ) (mod p) : mi , ni ≤ Cp log p, (mi ni , 23!) = 1 = Zp 2 4 by τ (12) = −τ (27) − τ (55) − τ (69) − τ (90) − τ (105). Then using the multiplicative property of τ (n), we conclude that the set {τ (n) (mod p) : n ≤ Cp2 log4 p} forms an additive basis of Zp of order at most 96. It is also useful to note that τ (6) = −6048, τ (14) = 401856, τ (29) = 128406630, τ (41) = 308120442, τ (42) = 101267712, τ (44) = −786948864, τ (48) = 248758272. Thus, we have a representation of zero by the sum of seven values of τ (n): τ (6) + τ (14) + τ (29) + τ (41) + τ (42) + τ (44) + τ (48) = 0. 3 Proof of Theorem 1 Let M be a large even integer parameter. Define the set Q = {q : 23 < q ≤ M 1/11 }. We call a subset Q′ ⊂ Q admissible if the equation 6 X τ (qi′ ) = i=1 12 X τ (qi′ ) i=7 ′ has no solutions in q1′ , . . . , q12 ∈ Q′ satisfying q1′ < . . . < q6′ ; ′ q7′ < . . . < q12 ; 6 ′ (q1′ , . . . , q6′ ) 6= (q7′ , . . . , q12 ). (2) Using different properties of τ (n) it is easy to check that there are admissible subsets with 12 elements. This can be derived if one combines the above mentioned congruences τ (q) ≡ 1 + q 11 τ (q) ≡ 1 + q 11 (mod 691), (mod 28 ), where q is an odd prime, with chinese remainder theorem and prime number theorem for arithmetical progressions to show that for any j, 1 ≤ j ≤ 12, one can find a sufficiently large prime ℓj satisfying τ (ℓj ) ≡ 2j (mod 8 × 691). Next, among all admissible subsets we take Q′ to be one with the biggest cardinality. If there are several such subsets, we take Q′ to be one of them. In particular, |Q′ | ≥ 12 and all the sums of the type 6 X τ (qi′ ) : i=1 q1′ < · · · < q6′ , q1′ , . . . , q6′ ∈ Q′ , are distinct. By Deligne’s estimate for τ (q) and pigeonhole principle we have |Q′ |6 ≪ (M 1/11 )11/2 , whence, |Q′ | ≪ M 1/11−1/132 . (3) Given q ∈ Q \ Q′ , consider the set Q′ ∪ {q}. From the maximality of |Q′ | we ′ know that there exist q1′ , . . . , q12 ∈ Q′ ∪ {q} such that 6 X τ (qi′ ) = i=1 12 X τ (qi′ ) (4) i=7 and q1′ < . . . < q6′ ; ′ q7′ < . . . < q12 ; ′ (q1′ , . . . , q6′ ) 6= (q7′ , . . . , q12 ). From the definition of Q′ we derive that ′ q ∈ {q1′ , . . . , q12 }. 7 (5) ′ Besides, due to (5), q occurs in the sequence q1′ , . . . , q12 at most two times. If q occurs there twice, then it occurs exactly one time in the sequence q1′ , . . . , q6′ and exactly one time in the sequence ′ q7′ , . . . , q12 . Therefore, we can cancel both sides of (4) by τ (q) and enumerating the remaining numbers qi′ , we obtain that there exist ′ ∈ Q′ q7′ , . . . , q11 q1′ , . . . , q5′ ∈ Q′ ; such that 5 X τ (qi′ ) = i=1 and q1′ < . . . < q5′ ; 11 X (6) τ (qi′ ) i=7 ′ q7′ < . . . < q11 ; ′ (q1′ , . . . , q5′ ) 6= (q7′ , . . . , q11 ). (7) Since |Q′ | > 10, there exists ′ q ′ ∈ Q′ \ {q1′ , . . . , q5′ , q7′ , . . . , q11 }. Then we have 5 X τ (qi′ ) ′ + τ (q ) = i=1 12 X (8) τ (qi′ ) + τ (q ′ ). i=7 This, in view of (6), (7) and (8), contradicts the definition of Q′ . Therefore, there is only the possibility that q occurs in the sequence ′ q1′ , . . . , q12 exactly one time. Thus, in view of (4), there exist q̃1 , . . . , q̃11 ∈ Q′ such that 6 X i=1 τ (q̃i ) = 11 X i=7 8 τ (q̃i ) + τ (q). Hence, for any q ∈ Q \ Q′ we get 6 X τ (q) i=1 6 X i=1 τ (q̃i q) − 11 X τ (q̃i q) − τ (q 2 ) = i=7 11 X τ (q̃i ) − τ (q) i=7 2 τ (q̃i ) − τ (q 2 ) = τ (q) − τ (q 2 ) = q 11 . Thus we have proved that for any element q ∈ Q \ Q′ there exist elements q̃1 , . . . , q̃11 ∈ Q′ such that q 11 = 6 X i=1 τ (q̃i q) − 11 X i=7 τ (q̃i q) − τ (q 2 ). (9) Our next aim is to prove the solubility of the Waring-Goldbach equation 2050 X qj11 = M (10) j=1 in primes q1 , . . . , q2050 ∈ Q \ Q′ . First of all, from (10) we have qj ≤ M 1/11 . Next, for the number of solutions of (10) with qj ∈ Q we have, according to Lemma 4 with s0 = 2050, the lower bound ≥ c1 M 2050/11−1 . (log M)2050 (11) For the number of solutions of (10) with at least one qj0 ∈ Q′ we have, according to Lemma 4 with s0 = 2049 and (3), the upper bound ≤ 2050c2 |Q′ | M 2050/11−1−1/132 M 2049/11−1 ≪ . (log M)2049 (log M)2049 (12) Thus, (11) > (12). Therefore, (10) is solvable in qj ∈ Q \ Q′ . We fix one of such solutions (q1 , . . . , q2050 ). To each qj , 1 ≤ j ≤ 2050, we apply (9) with q = qj and then perform the summation over 1 ≤ j ≤ 2050. Since q̃i q ≤ M 2/11 , we obtain M= 6×2050 X i=1 τ (ni ) − 9 6×2050 X i=1 τ (mi ), where max 1≤i≤6×2050 {ni , mi } ≤ M 2/11 , gcd(ni mi , 23!) = 1. Our assumption on M is that it is a large even integer. Clearly, we can multiply the above equality by −1 to have the same type of representation for −M as well. Furthermore, we can remove the parity condition on M by extracting one element τ (1) = 1 or τ (29) ≡ 0 (mod 2). Thus, any integer M with sufficiently large |M| can be represented in the form M= 6×2050 X i=1 τ (ni ) − 6×2050+1 X τ (mi ), (13) i=1 where max 1≤i≤6×2050+1 {ni , mi } ≤ |M|2/11 + 1, gcd(ni mi , 23!) = 1. Recall that (see (1)) −τ (12) = 370944 = τ (27) + τ (55) + τ (69) + τ (90) + τ (105). Therefore, multiplying (13) by −τ (12) and using the multiplicative property of τ (n), we obtain 6×6×2050+1 X 370944M = τ (ni ) (14) i=1 with max 1≤i≤6×6×2050+1 ni ≤ 106|M|2/11 . Let us show that any integer r, 0 ≤ r < 370944, can be expressed as a sum of say exactly 198 numbers of the form τ (n), n ≤ 105 (an extra effort would reduce 198 to a much smaller constant, which however do not essentially influence to our final result). To this end, we recall that τ (1) = 1, τ (2) = −24, τ (3) = 252, τ (5) = 4830, τ (8) = 84480. Now if 0 ≤ r < 370944, then r = 84480r5 + r4′ = τ (8)r5 + r4′ for some integers 0 ≤ r5 ≤ 4 and 0 ≤ r4′ < 84480. Next, any such r4′ can be expressed as r4′ = 4830r4 + r3′ = τ (5)r4 + r3′ , 10 where 0 ≤ r4 ≤ 17 and 0 ≤ r3′ < 4830. Any such r3′ can be written in the form r3′ = 252r3 − r2′ = τ (3)r3 − r2′ where 0 ≤ r3 ≤ 20 and 0 ≤ r2′ < 252. Any such r2′ can be written in the form r2′ = 24r2 − r1 = −τ (2)r2 − r1 , with 0 ≤ r2 ≤ 11 and 0 ≤ r1 ≤ 23. Thus we have r = τ (8)r5 + τ (5)r4 + τ (3)r3 + τ (2)r2 + τ (1)r1 . Therefore, to express such a given r, at most r5 + . . . + r1 ≤ 75 number of summands of τ (n), n ≤ 10, are sufficient. On the other hand, any integer greater than 29 can be expressed in the form 6x+7y with nonnegative integers x, y. Thus, in order to have a fixed number of summands for all r, we can use constructions (1) and (2). In particular, any integer r, 0 ≤ r < 370944, can be expressed in the form 198 X τ (ai ) i=1 with positive integers a1 , . . . , a198 ≤ 105. Let now N be an arbitrary integer with a sufficiently large modulo |N|. The above argument shows that for some positive integers a1 , . . . , a198 ≤ 105 we have 198 X N≡ τ (ai ) (mod 370944). i=1 Therefore, using (14), we obtain that N= 198 X τ (ai ) + 370944M = i=1 6×6×2050+199 X τ (ni ) = i=1 where max ni ≤ 106|M|2/11 ≤ 15|N|2/11 . 1≤i≤73999 11 73999 X i=1 τ (ni ), Thus, we have proved that there exists an absolute positive integer constant N0 such that for any integer N with |N| ≥ N0 the equation 73999 X τ (ni ) = N i=1 has a solution in positive integers n1 , . . . , n73999 ≪ |N|2/11 . Let now N be an arbitrary integer. If |N| > N0 then |N − τ (1)| ≥ N0 and therefore we can express the number N − τ (1) as a sum of 73999 values of τ (n) with n ≪ |N|2/11 . Theorem 1 follows in this case. If |N| ≤ N0 , then we take an integer constant n0 such that |τ (n0 )| > 2N0 . Then |N − τ (n0 )| > N0 . Thus N − τ (n0 ) can be expressed as a sum of 73999 values of τ (n), n ≪ 1. Theorem 1 is proved. Remark 1. One can easily see that the numbers ni constructed in the proof satisfy the condition τ (ni ) ≪ |N|. Remark 2. In our forthcoming paper we will prove that for any integer N with |N| ≥ 2 the diophantine equation 148000 X τ (ni ) = N i=1 has a solution in positive integers n1 , n2 , . . . , n148000 satisfying the condition max 1≤i≤148000 ni ≪ |N|2/11 e−c log |N |/ log log |N | , for some absolute constant c > 0. In view of Deligne’s estimate τ (n) ≤ n11/2 d(n), where d(n) is the number of divisors of n, this reflects the best possible bound for the size of the variables ni , apart from the value of the constant c. 12 4 Proof of Theorem 2 Let C be a large positive constant. Consider the sets Q = {q : 23 < q ≤ Cp1/2 log p} and I = {τ (q) (mod p) : q ∈ Q}. √ If |I| > 3 p, then we can split I into two subsets X , Y with |X ||Y| > 2p. The result in this case follows from Lemma 5. √ Let now |I| < 3 p. Then |I| [ Q= i=1 Ai , where the sets Ai are defined such that the condition q ′ , q ′′ ∈ Ai implies τ (q ′ ) ≡ τ (q ′′ ) (mod p). Clearly, for some A′i ⊂ Ai we have 0 ≤ |Ai| − |A′i| ≤ 3, Denote ′ Q = |A′i| ≡ 0 (mod 4). |I| [ i=1 A′i. By the prime number theorem we have |Q| ≥ Cp1/2 provided that p is large enough. Then |Q′ | ≥ |Q| − 3|I| ≥ |Q| − 9p1/2 ≥ (C − 9)p1/2 . (15) Since the cardinality of each set A′i is even, we can produce |A′i|/2 pairs formed with different primes of the set A′i . Then there are totally |I| X i=1 |A′i |/2 = |Q′ |/2 pairs (q, q ′ ). We divide this set of pairs into two disjoint subsets J1 and J2 , so that |J1 | = |J2 | = |Q′ |/4. Now consider the following two sets: X = {τ (qq ′ ) − τ (q 2 ) (mod p) : 13 (q, q ′ ) ∈ J1 } and Since Y = {τ (qq ′ ) − τ (q 2 ) (mod p) : (q, q ′) ∈ J2 }. τ (qq ′ ) − τ (q 2 ) = τ (q)τ (q ′ ) − τ (q 2 ) ≡ τ 2 (q) − τ (q 2 ) ≡ q 11 (mod p), and q 11 (mod p) can take any value at most 11 times, we have |X | ≥ |J1 |/11 = |Q′ |/44, |Y| ≥ |J2 |/11 ≥ |Q′ |/44. Therefore, if we choose C = 100 say, then according to (15), we obtain |X ||Y| > 2p. Applying Lemma 5, we finish the proof of Theorem 2. 5 Proof of Theorem 3 Consider the set of residue classes A′ = {τ (q) (mod p) : p/2 < q ≤ p}. From different properties of τ (n) it follows that A′ contains more than one element (apply, for example, the above mentioned congruence modulo 691 and the prime number theorem for arithmetical progressions). For a given a′ ∈ A′ , let I(a′ ) be the number of solutions of the congruence τ (q) ≡ a′ (mod p), p/2 < q ≤ p. From prime number theorem, X X I(a′ ) = 1 ≫ p log−1 p. a′ ∈A′ p/2<q≤p Therefore, there exists a′0 ∈ A′ such that I(a′0 ) ≫ p|A′ |−1 log−1 p. (16) Next, denote A = A′ \ {a′0 }. Since |A′| ≥ 2, then 0.5|A′ | ≤ |A| ≤ |A′|. (17) Now define the set B = {τ (q 2 ) (mod p) : p/2 < q ≤ p, τ (q) ≡ a′0 14 (mod p)}. The elements of the set B are of the form 2 τ (q 2 ) = τ 2 (q) − q 11 ≡ a′0 − q 11 (mod p) where p/2 < q ≤ p and τ (q) ≡ a′0 (mod p). Thus, according to (16) and (17), q runs through the set of ≫ p|A′|−1 log−1 p ≫ p|A|−1 log−1 p different residue classes modulo p. Since q 11 (mod p) can take any value at most 11 times, then |B| ≫ p|A|−1 log−1 p. (18) Fix ε, 0 < ε < 0.1. Let now C be the set of all different elements of the sequence of residues τ (q) (mod p), τ (q 2 ) (mod p), where q ≤ p0.5ε . The above argument applied to the sets {τ (q) (mod p) : q ≤ p0.5ε } {τ (q 2 ) (mod p) : q ≤ p0.5ε } and shows that |C| ≫ pε/6 . Note that the sets A, B, C are formed with elements of types τ (n1 ), τ (n2 ), τ (n3 ) correspondingly in such a way, that the numbers n1 , n2 , n3 are pairwise coprime and n1 ≤ p, n2 ≤ p2 , n3 ≤ pε . If |A| < p0.1ε , then by (18), |B| ≫ p1−ε/9 and thus |B||C| ≫ p1+0.01ε . Therefore we can apply Lemma 5 with X = B and Y = C and use the multiplicative property of τ (n). The elements of the set X Y = {xy : x ∈ X , y ∈ Y} in this case will be of the form τ (n) (mod p) with n ≤ p2+ε . If |A| > p2/3 , then we first split A into two disjoint subsets A1 and A2 with ≥ p2/3 /3 elements in each. Then apply Lemma 5 with X = A1 and Y = A2 . The elements of the set X Y in this case take the form τ (n) (mod p) with n ≤ p2 . 15 If p0.1ε ≤ A ≤ p2/3 , then denote by T the one of the sets A + C and AC with the biggest cardinality. Since |C| ≫ pε/6 , according to Bourgain’s estimate [2, Theorem 1.1], there exists a positive constant γ = γ(ε) > 0 such that |T | ≫ pγ |A|. Then |B||T | ≫ p1+γ/2 . Hence, we can apply Lemma 5 with X = B and Y = T . The elements of the set X Y in this case will be either of the form τ (n1 ) + τ (n2 ) (mod p) with n ≤ p3 or of the form τ (n) (mod p) with n ≤ p3+ε . The result now follows. References [1] G. I. Arkhipov and V. N. Chubarikov, O chisle slagayemix v additivnoy probleme Vinogradova i ee obobsheniyax, IV Mezhdunarodnaya konferentsiya “Sovremenniye problemy teorii chisel i ee prilozheniya”, Aktualniye problemy, Chast I, 5–38 (2001). [2] J. Bourgain, More on the sum-product phenomenon in prime fields and its applications, Int. J. Number Theory, 1 (2005), 1–32. [3] J. Bourgain, N. Katz and T. Tao, A sum-product estimate in finite fileds and their applications, Geom. Func. Anal., 14 (2004), 27–57. [4] P. Deligne, La conjecture de Weil I, (French) Inst. Hautes tudes Sci. Publ. 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Swinnerton-Dyer] (French), Lecture Notes in Math, 317, 319–338, Springer, Berlin, 1973. [13] I. E. Shparlinski, On the value set of the Ramanujan function, Arch. Math., 85 (2005), 508–513. Adresses of the authors: M. Z. Garaev and V. C. Garcia: Instituto de Matemáticas UNAM, C.P. 58089, Morelia, Michoacán, México. S. V. Konyagin: Dept. of Mechanics and Mathematics, Moscow State University, Moscow, 119992, Russia. emails: [email protected] [email protected] [email protected] 17