aa r X i v : . [ m a t h . N T ] A ug SOME DIOPHANTINE EQUATIONS ANDINEQUALITIES WITH PRIMES
ROGER BAKER
Abstract.
We consider the solutions to the inequality | p c + · · · + p cs − R | < R − η (where c > c N and η is a small positive number; R is large).We obtain new ranges of c for which this has many solutions inprimes p , . . . , p s , for s = 2 (and ‘almost all’ R ), s = 3, 4 and 5.We also consider the solutions to the equation in integer parts[ p c ] + · · · + [ p cs ] = r where r is large. Again c > c N . We obtain new ranges of c for which this has many solutions in primes, for s = 3 and 5. Introduction
Let c > c N . Let η be a small positive number depending on c .Let R be a large positive number. We consider solutions in primes ofthe inequality(1) s | p c + · · · + p cs − R | < R − η first studied by ˘Sapiro-Pyateck ( ıı [38]. We also consider the equation ininteger parts(2) s [ p c ] + · · · + [ p cs ] = r. We give results providing large numbers of solutions of (1) s ( s =2 , , ,
5) and (2) s ( s = 3 ,
5) for new ranges of c . For s = 2, oneneeds to restrict R to ‘almost all’ real numbers in an interval [ V, V ].Following a nice innovation in a paper of Cai [10], there has beenrecent progress in all these cases; see below for details. In the presentpaper progress is made by combining this innovation with the powerful Mathematics Subject Classification.
Primary 11N36; secondary 11L20,11P55.
Key words and phrases. exponential sums, the alternative sieve, the Hardy-Littlewood method, the Davenport-Heilbronn method. exponential sum bounds of Huxley [23], Bourgain [6], and Heath-Brown[22]. When discussing (1) , (1) and (2) , we use a vector sieve inconjunction with the Harman sieve. The other cases are simpler andHeath-Brown’s generalized Vaughan identity replaces the sieve method.We write ‘ n ∼ N ’ to signify N < n ≤ N . Let X = R /c .Let A s ( R ) denote the number of solutions of (1) s with X < p j ≤ X ( j = 1 , . . . , s ). Let B s ( r ) denote the number of solutions of (2) s with X < p j ≤ X ( j = 1 , . . . , s ). One expects heuristically to obtain (atleast for c not too large) the bounds(3) s A s ( R ) ≫ R sc − − η (log R ) s and(4) s B s ( r ) ≫ r sc − (log r ) s . Theorem 1.
Let V be large. Suppose that c < = 1 . . . . , c = 4 / .We have (3) for all R in [ V, V ] except for a set of R having measure O ( V exp( − C (log V ) / )) . (We denote by C a positive absolute constant, not the same at eachoccurrence.)Previous upper bounds for permissible c :17/16 [26], 15/14 [27], 43/36 [16], 59/44 = 1.3409. . . [12]. Theorem 2.
Let R be large. Suppose that c < / . Then (3) holds. Previous upper bounds:15/14 [26], 13/12 [8], 11/10 [9, 25], 237/214 [13],6155 [24], 109 [5], 4336 = 1 . . . . [10]. Theorem 3.
Let R be large. Suppose that c < / . Then (3) holds. Previous upper bounds:9781 [35], 65 [31], 5944 [12], 1198889 = 1 . . . . [33]. Theorem 4.
Let R be large. Suppose that c < = 2 . . . . . Then (3) holds. OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 3
Previous upper bounds:1 . . . . [14], 1 + √
52 [18], 8140 [15], 10853 [35], 2.041 [4],2.08 [12], 665576319965 = 2 . . . . [30]. Theorem 5.
Let n be large. Suppose that c < = 1 . . . . . Then (4) holds. Previous upper bounds:1716 [28], 1211 [25], 258235 [13], 137119 [11], 31132703 = 1 . . . . [32]. Theorem 6.
Let n be large. Suppose that c < = 2 . . . . . Then (4) holds. Previous upper bounds:41090541999527 [29], 408197 = 2 . . . . [34].Along usual lines, we employ a continuous function φ : R → [0 , φ ( y ) = 0 ( | y | ≤ R − η ) , φ ( y ) = 1 (cid:18) | y | ≤ R − η (cid:19) , with Fourier transformΦ( x ) := Z ∞−∞ e ( − xy ) φ ( y ) dy , where e ( θ ) := e πiθ , satisfying(1.2) Z | x | >X η | Φ( x ) | dx ≪ X − . We define τ = X η − c , K = X η , L = log X, P ( z ) = Y p Let ρ ( n ) denote the indicator function of the prime numbers. For u ∈ N , z > 1, let ρ ( u, z ) = 1 if ( u, P ( z )) = 1 . Let ρ ( u, z ) = 0 otherwise. For a vector sieve one usually uses functions ρ − ( . . . ), ρ + ( . . . ) with ρ − ( n ) ≤ ρ ( n ) ≤ ρ + ( n );but (without loss) we shall take ρ − = ρ , so that the inequality basicto [7] becomes(1.3) ρ ( m ) ρ ( ℓ ) ≥ ρ + ( m ) ρ ( ℓ ) + ρ ( m ) ρ + ( ℓ ) − ρ + ( m ) ρ + ( ℓ ) . We shall need exponential sums S ( x ) = X X We now describe briefly the underlying principle of the proofs. For(3) , (3) we use L s A s ( R ) ≫ X X
For N ≥ 1, we write I ( N ) for a subinterval of ( N, N ], not the sameat each occurrence. For a real function f on [ N, N ] we write S ( f, N ) = X n ∈ I ( N ) e ( f ( n )) . The fractional part of x is written as { x } . We write A ≍ B for A ≪ B ≪ A . Implied constants in the conclusions of the lemmas depend OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 7 on c , η if these appear, together with any implied constants in thehypotheses. Lemma 1. Let x > . For real numbers a n , | a n | ≤ , let W ( X, x ) = X X Summing over n , W ( X, x ) = X X Lemma 3 ( A process) . For ≤ Q ≤ N , S ( f, N ) ≪ NQ X | q | Define x ν by f ′ ( x ν ) = ν and let φ ( ν ) = − f ( x ν ) + νx ν . Then for I ( N ) = [ a, b ] , S ( f, N ) = X f ′ ( b ) ≤ ν ≤ f ′ ( a ) e ( − φ ( ν ) − / | f ′′ ( x ν ) | / + O (log( F N − + 2) + F − / N ) . Proof. [19, Lemma 3.6]. (cid:3) Lemma 5. (i) Let ℓ ≥ be a given integer, L = 2 ℓ . Suppose that f has ℓ + 2 continuous derivatives on [ N, N ] and | f ( r ) ( x ) | ≍ F N − r ( r = 1 , . . . , ℓ + 2 , x ∈ [ N, N ]) . Then S ( f, N ) ≪ F / (4 L − N − ( ℓ +2) / (4 L − + F − N. (ii) Let f ( x ) = yx b where y = 0 , bb − 6∈ { , , . . . , ℓ + 1 } . With F = | y | N c and ℓ , L as in (i), S ( f, N ) ≪ F / − ℓ +14 L − N ℓ +24 L − + N F − . Proof. Part (i) is Theorem 29 of [19]. If F N − < η , part (ii) followsfrom the Kusmin-Landau theorem. Suppose now that F N − ≥ η . Weapply the B process to S ( f, N ) and then part (i) of the lemma to theresulting sum (after a partial summation). This yields S ( f, N ) ≪ N F − / (cid:16) F / (4 L − ( F N − ) − ℓ +24 L − + ( F N − ) F − (cid:17) + N F − / + 1 . The last three terms can be absorbed into the first term. (cid:3) Lemma 6. Suppose that g (6) is continuous on [1 , and g ( j ) ( x ) ≪ ≤ j ≤ g ( j ) ( x ) ≍ ≤ j ≤ . Let T , N be positive with T / ≪ N ≪ T / . Let S h = X m ∈ I h e (cid:16) T y h g (cid:16) mN (cid:17)(cid:17) where I h is a subinterval of [ N, N ] and y , . . . , y H ∈ [1 , with y j +1 − y j ≫ H ( j = 1 , . . . , H − . Then H X h =1 | S h | ≪ H / N / T + η + HN / T / η . Proof. We combine a special case of [23, Theorem 2] with H¨older’sinequality. (cid:3) Lemma 7. Let g be a function with derivatives of all orders on (cid:2) , (cid:3) and | g ( j ) ( x ) | ≫ (cid:18) x ∈ (cid:20) , (cid:21) , ≤ j ≤ (cid:19) . Let T , N be positive with T / ≤ N ≤ T / . Then X n ∈ I ( N ) e (cid:16) T g (cid:16) nN (cid:17)(cid:17) ≪ N / T + η . Proof. Theorem 4 of [6] is the case T / ≤ N ≤ T , I ( N ) = [ N, N ] . On pages 222–223 of [6] it is indicated how to extend this to h T , T / i using the B process. An application of [19, Lemma 7.3] enables one toreplace [ N, N ] by I ( N ) with the loss of a log factor. (cid:3) Lemma 8. Let k ∈ N , k ≥ . Let f have continuous derivatives f ( j ) (1 ≤ j ≤ k ) on [0 , N ] , | f ( k ) ( x ) | ≍ λ k on (0 , N ] . OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 11 Then (2.1) S ( f, N ) ≪ N η (cid:16) λ k ( k − k + N − k ( k − + N − k ( k − λ − k k − k (cid:17) . Proof. [22, Theorem 1]. (cid:3) Lemma 9. Let θ , φ be real constants, θ ( θ − θ − φ ( φ − θ + φ − θ + φ − θ + 2 φ − θ + φ − = 0 . Let F ≥ and let | a m | ≤ . Let T ( θ, γ ) = X m ∼ M a m X n ∈ I m e (cid:18) F m θ n φ M θ N φ (cid:19) where I m is a subinterval of ( N, N ] . Then T ( θ, γ ) ≪ ( M N ) η ( F / M / N / + F / M / N / + F / M / N / + M / N + M N / + M N F − ) . Proof. [5, Theorem 2]. (cid:3) We write ( α ) = 1, ( α ) s = ( α ) s − ( α + s − 1) ( s = 1 , , . . . ). Lemma 10. Let θ , φ be real ( θ ) ( φ ) ( θ + φ + 2) = 0 . Let M N ≍ X , F ≥ , | a m | ≤ . Let N = min( M, N ) . Then in thenotation of Lemma 9, T ( − θ, − φ ) ≪ X η ( X / + XN − / + F / X / N − / + ( F X N − N − ) / + XF − ) . Proof. [4, Lemma 9]. (cid:3) Lemma 11. Let | a m | ≤ , | b n | ≤ . Let S = X m ∼ M a m X n ∼ N b n e ( Bm β n α ) where M ≥ , N ≥ , α ( α − α − β ( β − β − = 0 . Supposethat F := BM β N α ≫ X. Then SX − η ≪ F / N / M / + F / N / M / (2.2) + F / N / M / + F / N / M / + F / N / M / + F / N / M / + F / N / M / + N / M + F / ( N M ) / . Proof. This is due to Sargos and Wu [39]. Full details are given in [5,proof of Theorem 3]. (cid:3) Lemma 12. Let < B < K and | c n | ≤ . Let V ( x ) = X X 0) [( n + j ) c ] − [ n c ] = ( n + j ) c − n c + O (1) ≍ jn c − .(ii) Again, we give details only for (2.3). The left-hand side in (ii) is OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 13 X X The following result abstracts the idea of Cai mentioned in Section1. Lemma 13. Let µ be a complex Borel measure on [ X − c , K ] . Let λ , . . . , λ N ∈ R . Let a n (cid:0) X < n ≤ X (cid:1) be real numbers, S ( x ) = X X Let G be a complex function on [ X, X ] . Let u ≥ , v , z be numbers satisfying u ≤ z , uz ≤ X and X ≤ v . Then X X Lemma 15. Let w ( . . . ) be a complex function with support on (cid:2) X , X (cid:3) ∩ Z , | w ( n ) | ≤ X /η for all n . For m ∈ N , z ≥ let S ( m, z ) = X ( n, P ( z ))=1 w ( mn ) . Let α > , < β ≤ / , M ≥ , Y > . Suppose that whenever | a m | ≤ , | b n | ≤ d ( n ) , we have X m ≤ M a m X n w ( mn ) ≪ Y, (2.6) X X α ≤ m ≤ X α + β a m X n b n w ( mn ) ≪ Y. (2.7) Let | u ℓ | ≤ , | v s | ≤ , for ℓ ≤ R , s ≤ S , also u ℓ = 0 for ( ℓ, P ( X η )) > , v s = 0 for ( s, P ( X η )) > . Suppose that R < X α , S < M X − α . Then X ℓ ≤ R X s ≤ S u ℓ v s S ( ℓs, X β ) ≪ Y L . Proof. [5, Lemma 14]. (cid:3) Lemma 16. Let α > , < β ≤ / , Y > , y > . Suppose thatwhenever | a m | ≤ , | b n | ≤ we have S := X X α ≤ m ≤ X α + β X X Let | a m | ≤ , | b n | ≤ . Let S = X ∗ p ,...,p s X a m X b nX α ≤ mp ...p r ≤ X α + βX Let c < . and x ∈ ( τ, X − c ] . Let V ( x ) = X m ∼ M X n ∼ N X In each case we use Lemma 4 with ℓ = 2. Treating the simpler bounds(3.7), (3.8) first, we bound the sum over n in (3.7), (3.8) by ≪ F / X / + F − X ≪ h / X / , which yields (3.7), (3.8).For (3.6), take Q = ηN . Arguing as in [1, proof of Theorem 5], wehave(3.9) | S ( γ ) | ≪ X Q + XQ X q ≤ Q X n ≪ N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X m ∈ I ( M ) e (( h + γ ) m c (( n + q ) c − n c ))) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . For the inner sum on the right-hand side we use Lemma 4 with ℓ = 2,obtaining the bound ≪ ( qN − HX c ) M + M | γ | qN − X c . Thus | S ( γ ) | ≪ X N + XN ( HX c ) + X − η since | γ | X c > X η . This gives the desired bound (3.5).For the case b n = 1 identically, it suffices to add the bound(3.10) X n ∈ I ( N ) e (( h + γ ) m c n c ) ≪ N X − η whenever N ≥ X − η . We bound the left-hand side by ≪ ( HX c ) N + ( xX c ) − N and obtain (3.10) at once.We now deduce (3.1) for 2 ≤ s ≤ 4. We find that Lemma 17 yields(3.11) U ( x ) ≪ X − η ( τ < x < X − c )(here we require the reader to look ahead to the form of S + ( x ) and T + ( x ) in later sections, or else use Heath-Brown’s identity if appro-priate). Now (recalling Lemma 12(i)) a simple splitting-up argumentyields (for some B < X η ) Z X − c τ | U ( x ) | s Φ( x ) dx ≪ L X − cη sup y ∈ [ τ,K ] | U ( y ) | s − Z Kτ | U ( x ) | dx ≪ L X − cη X s − − η X − c + η ≪ X s − c − η . The argument for (3.2) is very similar using Lemma 12 (ii). (cid:3) Minor arc in Theorems 1 and 3 We shall show that(4.1) Z KX − c | S ( x ) | dx ≪ X − c − η . Since Φ( x ) ≪ X − cη , this reduces the integral in (1.4), in effect, to themajor arc in Theorem 3. For Theorem 1, let E ( x ) = ( S ( x )Φ( x ) x ∈ [ τ, K ]0 otherwise.By Parseval’s formula, Z VV | b E ( R ) | dR < Z Kτ | E ( x ) | dx ≪ ( V /c ) − c − cη . Hence b E ( R ) = Z R S ( x )Φ( x ) e ( − Rx ) dx satisfies | b E ( R ) | < V c − − η except for a set of measure O ( V − η ) in [ V, V ]. Again, this gives thedesired ‘reduction to the major arc.’To prove (4.1) we first apply Lemma 13(ii) with S ( x ) = S ( x ) , J ( x ) = A ( x ) , G ( x ) = ¯ S ( x ) S ( x ) . From Lemma 12 and the Cauchy-Schwarz inequality, Z Kτ | G ( y ) | dy ≪ L (cid:18)Z BB | S ( x ) | dx (cid:19) / (cid:18)Z BB | S ( x ) | dx (cid:19) / OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 21 (for some B ∈ [ τ, K ]) ≪ X − c +2 η . We shall show below that (2.4) holds with(4.2) U = X − c − η and that(4.3) S ( x ) ≪ X (7 − c ) / − η . Hence Z KX − c | G ( y ) | dy ≪ X (7 − c ) / − η Z KX − c | G ( y ) | dy ≪ X − c − η . Now Lemma 13(ii) yields (cid:18)Z Kτ | S ( x ) | dx (cid:19) ≪ X − c +6 − c − η + X − c − η +6 − c +2 η ≪ X − c − η as required for (4.1).Turning to the proof of (4.2), we apply the B process first, followedby a partial summation and then Lemma 8 with k = 5. This is legiti-mate since the B process produces a sum of the form X n ∈ I e ( yn c/ ( c − )where yn c − ≍ F := xX c ;and the five differentiations required in Lemma 8 are permissible unless(4.4) cc − m ∈ N , m ≤ . We have excluded c = 4 / 3, so (4.4) cannot hold. The error term inLemma 4 is ≪ L + F − / X ≪ X / (since x ≥ X − c ), which is acceptable. We may take U ≪ XF − N η n ( F N − ) + N − + ( F N − ) − N − o where N = F X − . Here XF − N η ( F N − ) ≪ X + η F / ≪ X − c − η since c < < . Next, XF − / N + η ≪ X c +120 +2 η ≪ X − c − η since c < . Finally XF − ( F N − ) − N + η ≪ X η F / ≪ X − c − η since c < < .We now use Lemma 14 to prove (4.3). Here and below, we take G ( n ) = e ( xn c ) in Lemma 14. A Type I sum will be of the form S I ( x ) = X m ∼ M a m X n ∼ N X 1. Taking v = X . , u = X . , z = X . in Lemma 14, it suffices to show that S I ( x ) ≪ X (7 − c ) / − η for N ≥ z (4.5)and S II ( x ) ≪ X (7 − c ) / − η for u ≤ N ≤ v. (4.6) OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 23 For (4.5), we appeal to Lemma 10. We have (7 − c ) / > . S I ( x ) are acceptable, while XF − ≪ . Next, for N ≥ z , F / X / N − / ≪ X . / / − . / ≪ X . . Finally, we have a term that is ≪ ( F X ) / + ( F X N − ) / ≪ X (1 . / + X (1 . − . / ≪ X . . For (4.6), we apply the obvious variant of (3.9), taking Q = X . to give an acceptable term X /Q . It remains to show that for q ≤ Q , n ≤ N ,(4.7) S n,q := X m ∼ M e ( xm c (( n + q ) c − n c )) ≪ M Q − . Here F is replaced by F := xqX c N − . We apply Lemma 5 (ii) with ℓ = 4 to obtain S n,q ≪ F / M / ≪ M X − . since X / N − / > X . > X (39 / . / . . However, the sixdifferentiations are only permissible when c = 1 + 1 m where m ≤ 4. We excluded m = 3, so we now need to treat c = sep-arately. Here we use Lemma 5 (ii) with ℓ = 3; the five differentiationsare permissible and S n,q ≪ F / M / ≪ M X . by a similar calculation. This completes the discussion of the minorarc. 5. Minor arc in Theorem 2. We shall set up a suitable function ρ + based on Type I and Type IIinformation. To obtain a negligible contribution of the minor arc we require(5.1) Z KX − c S ( x ) G ( x ) dx ≪ X − c − η for the two functions(5.2) G ( x ) = S + ( x ) S ( x )Φ( x ) e ( − Rx ) , S + ( x ) Φ( x ) e ( − Rx ) . It will suffice to show that A ( x ) ≪ X c + η + X − c | x | − ( | x | < X η )(5.3)and S + ( x ) ≪ X − c − η ( X − c ≤ x ≤ K ) . (5.4)We then apply Lemma 13 (ii) with S ( x ) = S ( x ), T ( x ) = A ( x ), G ( x )as in (5.2) so that Z KX − c | G ( x ) | dx ≪ X η and, using (5.4), Z KX − c | G ( x ) | dx ≪ X − c − η Z KX − c | G ( x ) | dx ≪ X − c − η . Thus (cid:18)Z KX − c S ( x ) G ( x ) dx (cid:19) ≪ L X − c X − c − η + L X c +1+ η +2(1+2 η ) ≪ X − c − η using c < / 5, which proves (5.1).To obtain (5.3) we use the Kusmin-Landau theorem if X c − | x | < η .Otherwise, we use the B process, giving a main term ≪ F ≪ X c + η OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 25 where F = xX c , and error terms ≪ F − / X + L ≪ X / . Aiming towards the definition of S + ( x ), we claim that Type II sumsare ≪ X . for either of the alternatives X / ≪ N ≪ X / (5.5)and X / ≪ N ≪ X / . (5.6)For (5.5) we begin with (3.9), replacing h + β by x , and taking Q = X . ≪ N . It remains to show that for given Q ∈ (cid:2) , Q (cid:3) and n ∼ N ,we have S ∗ := X q ∼ Q X n ∈ I ( M ) e ( x (( n + q ) c − n c ) m c ) ≪ Q M X − / . Following the analysis on pp. 171–172 of [5], we find that for some q ∼ Q , and R at our disposal with R ≪ N , and some r ∼ R , we have(5.7) S ∗ L ≪ N M F Q + N M + X N Q F Q (cid:18) N R + N | S ( n, q, r ) | (cid:19) . Here N ≍ F Q /X ≪ X . , t ( n, q ) = ( n + q ) c − ( n − q ) c ,t ( n , r ) = ( n + r ) c/ ( c − − ( n − r ) c/ ( c − , and we define S ( n, q, r ) = X n ∈ I ( r ) e (cid:16) C ( xX c t ( n, q )) − c t ( n , r ) (cid:17) with I ( r ) a subinterval of [ N , N ]. We choose R so that M N Q N F Q R = X Q , that is, R = N Q N F ≍ F N Q X . We have R ≪ N since N Q ≪ N Q ≪ X . The terms N M /F Q and N M in (5.8) are ≪ X /Q since N M F Q Q X = N QF ≪ N QX ≪ ,N M Q X ≪ Q M ≪ . For S ( n, q, r ), it suffices to show that S ( n, q, r ) ≪ X . /N. For then X F Q N Q N S ( n, q, r ) ≪ X F Q X . Q F Q X ≪ X Q . We apply Lemma 7 to S ( n, q, r ) with (taking 2 η < . − c ). We have T ≍ F Q N rN ≍ XrN ; T ≪ XRN ≍ F Q X ≪ X . − η . Provided that T / ≤ N ≤ T / , we obtain S ( n, q, r ) ≪ X η T / N / ≪ X η + + − η ≪ X . N ( N ≪ X / ) . We certainly have N ≤ T / since X . < (cid:18) XN (cid:19) / X − η as N < X . < X − . × − η . We may have N < T / . In this case we apply Lemma 4 with ℓ = 2to S ( n, q, r ). The term T − N is ≪ 1, so that S ( n, q, r ) ≪ T / N / ≪ T + < X . N OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 27 since T < X / , N < X / . This completes the proof that S II ≪ X . when (5.5) holds.Now suppose that (5.6) holds. We apply Lemma 11. Five of theterms U , U , . . . , U on the right-hand side of (2.2) are acceptable for N ≪ X . : U ≪ F / N / X / , . 220 + 9 / 240 + 2940 = 0 . . . . ,U ≪ F / N / X / , . 646 + 11 / 246 + 3246 = 0 . . . . ,U ≪ F / N / X / , . 210 + 3 / 210 + 35 = 0 . ,U ≪ F / N / X / , . 211 + 1 / 233 + 1722 = 0 . . . . ,U ≪ F / N / X / , . 25 + 1 / 210 + 35 = 0 . . Also U ≪ XN − / ≪ X . for N > X . . For the remaining terms, U ≪ F / N / X / ≪ X . for N ≪ X . ,U = F / X / N − / ≪ X . for N ≫ X / , while the bound U = F / X / ≪ X . ( F ≪ X / )actually determines our range of c .Finally we consider Type I sums, using S I ≪ M F / N / , which follows from Lemma 4 with ℓ = 2. Here M F / N / ≪ X + . M ≪ X . for M ≪ X . . If X . ≪ M ≪ X . we treat S I as a Type II sum. Hence S I ≪ X . for M ≪ X . . We now apply Lemma 15, taking w ( n ) = e ( xn c ), α = and β = − = , M = X . , S = 1. Thus X ℓ ≤ X / u ℓ S ( ℓ, X β ) ≪ X . L for any coefficients u ℓ with | u ℓ | ≤ u ℓ = 0 for ( ℓ, P ( X η )) > 1. (For X / < ℓ ≤ X this uses Lemma 16.) We use Buchstab’s identity ρ ( u, z ) = ρ ( u, w ) − X w ≤ p 12 + log 32 log X (cid:19) . Let S j ( x ) = X α ∈ I j X X
OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 29 (5.9) S ( x ) = S ( x ) − X j =1 S j ( x ) − D ( x ) . We use Buchstab’s identity again for S ( x ): S ( x ) = X α ∈ I X X α ≤ α + α < log 2 X log X e ( x ( p p ) c ) , S ( x ) = X α ∈ I X α ≥ α >α ≤ α + α + α < log 2 X log X e ( x ( p p p ) c ) . Finally,(5.14) S ( x ) = D ( x ) + K ( x ) , where D ( x ) is the part of the sum defining S ( x ) for which α + α ≤ + log 2log X , and K := S − D . Combining (5.9)–(5.14), ourdecomposition of S ( x ) is S = S − S + S − D − K − S − D − D − K − S − D . We define S + = S − S + S − K − S − K − S (5.15) = S + X j =1 D j . We observe firstly that S + ( x ) = X X 5. Hence α + α ∈ I ∪ I . Similarly for α + α .(ii) If β ≤ α < α < α < / α + α ∈ I , α + α ∈ I , α + α + α I , then α + α + α ≤ · = , hence α + α + α < , while α + α + α ≥ · > ; α + α + α ∈ I .(iii) If α < α ≤ α , ≤ α < , 1 ≤ α + α + α < log 2 X log X , and α + α ≥ + log 2log X , then α ≤ . Moreover α ≥ ( α + α + α ) ≥ , hence α ∈ I .In Section 9 we shall quantify the contribution of the functions D j ( x )to the integral in (1.6); similarly for Theorem ?? . OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 31 Minor arc in Theorem 5. In the present section we show that for c < , we have(6.1) Z X − c T ( x ) G ( x ) dx ≪ X − c − η , where G ( x ) = T ( x )Φ( x ) e ( − Rx ) . We apply Lemma 13 (ii) with S ( x ) = T ( x ), J ( x ) = B ( x ). To prove(6.1) it suffices to show that B ( x ) ≪ X − c − η + L X − c | x | − (0 < | x | < X η )(6.2)and T ( x ) ≪ X − c − η (cid:18) X − c ≤ x < (cid:19) . (6.3)For then Lemma 13 (ii) (with G ( x ) = 0 for x > ) yields (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z X − c T ( x ) G ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≪ X − c L max x ∈ [ X − c , ] | T ( x ) | Z X − c | T ( x ) | dx + X − c − η L Z X − c | T ( x ) | dx ! . The first summand on the right-hand side is ≪ X − c +3 − c − η +1+3 η ≪ X − c − η by (6.3) and Lemma 12 (i). The second summand is ≪ X − c − η X η ) ≪ X − c − η by (6.2) and Lemma 12 (i).For (6.2), we use Lemma 1 with a n = 1. We take H = X c − η . Since { x } = x in the sum, our objective is to show that we have X ≤ h ≤ H min (cid:18) , h (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X X For (6.13), it suffices to show that(6.15) X m ∼ M X ℓ ∼ N X M Q − , as we easily verify. Since qN − x X c ≥ XN − , the second term is acceptable.For (6.14) it suffices with the same λ to show that X m ∼ M X n ∼ N X Here we use (1.6), so in the present section we show that (with S + to be specified below)(7.1) Z KX − c S ( x ) G ( x ) dx ≪ X − c − η where G ( x ) is either S ( x ) S + ( x )Φ( x ) e ( − Rx ) or S ( x ) S + ( x ) Φ( x ) e ( − Rx ).Let us write k . . . k for sup norm on [ X − c , K ]. It suffices to showthat(7.2) A ( x ) ≪ X − c − η + X − c x − (0 < x ≤ X η )and that k S k ∞ ≪ X / η ;(7.3) k S + k ∞ ≪ X . η . (7.4)Using the bounds in Lemma 13 (ii), together with Lemma 12 (ii), (cid:12)(cid:12)(cid:12)(cid:12)Z KX − c S ( x ) G ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) ≪ L X − c ( k S k ∞ k S + k ∞ + k S + k ∞ ) X η + X − c − η X L X η ≪ X − c +2 ( +0 . ) +2+4 η + X − c − η ≪ X − c − η as required for (7.1).We turn to (7.2). This is obtained from Lemma 7 with xX c , X inplace of T , N . The Kusmin-Landau theorem gives A ( x ) ≪ X − c | x | − unless X c − x ≫ 1, which we now assume. If X ≤ ( xX c ) / we can use Lemma 7, since( xX c ) / ≪ X . × / ≪ X − η ; OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 37 we obtain A ( x ) ≪ ( X c +2 η ) + η X ≪ X − c − η since 13 c + 42 < − c (this inequality determines the range of c in the theorem).In the remaining case X > ( xX c ) / , we apply Lemma 5 (i) with ℓ = 1: A ( x ) ≪ ( xX c ) / X ≪ X / + η = X . η which suffices for (7.2).Turning to (7.3), we first show that Type II sums are O ( X / η )whenever X / ≪ N ≪ X (and hence whenever X / ≪ N ≪ X / ). Proceeding as in (3.9),we need to show X m ∼ MX As for (7.4), we take(7.5) S + ( x ) = X X We may now apply Lemma 15 with α = 0 . β = 0 . M = X . , R = S = 1, w ( n ) = e ( xn c ). We obtain the desired bound X n ∼ X ρ ( n, X . ) e ( xn c ) ≪ X . η , while the sum X X . ≤ p ≤ X . X X
We shall show, for suitably chosen T + ( x ), that(8.1) Z / X − c T ( x ) G ( x ) dx ≪ X − c − η where G ( x ) is either of T ( x ) T + ( x )Φ( x ) e ( − Rx ) or T ( x ) T + ( x ) Φ( x ) e ( − Rx ).In Lemma 13 (ii) we take S ( x ) = T ( x ) , J ( x ) = B ( x )and G as above. Suppose for the moment that(8.2) B ( x ) ≪ X − c − η + L X − c x − (0 < x ≤ K )and that, with T + ( x ) = X X 4) + 449690 + 63 c < − c, which reduces to c < . (This determines the range of c in Theorem5). In (8.8) we require12 + (3 c − 4) 141950 < − c, which holds for c < with a little to spare. OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 41 The contribution to the left-hand side of (8.6) from h ∼ H , H ≤ H < X − c , can be handled using (8.7), (8.8) since we have(2 H X c ) ≤ X ≤ ( H X c ) . The additional factor H/H arising from H/h leads to a negativeexponent of H in using (8.7), (8.8).For H ≥ X − c , we use Lemma 5 (i) with ℓ = 2: we need to verifythat HH ( H X c ) X < X − c − η . The worst case is H = X − c . Here H H X c X < H X − c − η since 70 c < T ( x ) and T + ( x ) using Lemma 1. For T ( x ), we choose H = X / . Now for (8.3) it suffices to show that for X / ≪ N ≪ X / , X < X ′ ≤ X , 1 ≤ h ≤ H , we have(8.9) S II := X m ∼ M a m X n ∼ N X The arguments in the present section are adapted from [4, 24, 28].We begin with a number of lemmas. Let OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 43 v ( X, x ) = Z X e ( xγ c ) dγ,v ( X, x ) = X ≤ m ≤ X c m /c − e ( xm ) . Proofs of Lemmas 18 and 19 (ii) can be found in Vaughan [41, Sections2.4, 2.5] with the unimportant difference that c ∈ N in [41], whileLemma 19 (i) follows from [19, Lemma 3.1]. Lemma 18. We have v ( X, x ) = v ( X, x ) + O (1 + X c | x | ) . Lemma 19. (i) We have v ( X, x ) − v ( X/ , x ) ≪ ( | x | X c − ) − . (ii) For | x | ≤ / , we have v ( X, x ) ≪ | x | − /c . Lemma 20. For ≤ s ≤ and r large, let X = r /c . We have L s := Z / − / (cid:18) v ( X c , x ) − v (cid:18)(cid:18) X (cid:19) c , x (cid:19)(cid:19) s e ( − rx ) dx ≫ r s/c − . Proof. The integral is1 c s X ( X ) c For ≤ s ≤ , we have H s = Z XX/ · · · Z XX/ φ ( t c + · · · + t cs − R ) dt . . . dt s ≫ X s − c − cη . Proof. One verifies easily that for each choice of t , . . . , t s − from (cid:2) X , X (cid:3) ,there is an interval of t s in [ X, X ] of length ≫ X − c − cη on which φ ( t c + · · · + t cs − R ) = 1 . (cid:3) A polytope means a bounded intersection of half-spaces in R j . Thepolytope P j is defined by P j = (cid:26) ( y , . . . , y j ) : β ≤ y j < y j − < · · · < y ,y + · · · + y j − + 2 y j ≤ L (cid:27) , where β = 8 / 75. In writing sums containing p , . . . , p j , it is convenientto set α j = ( α , . . . , α j ) := 1 L (log p , . . . , log p j ) ,f ( α ) = α − , f j ( α j ) = ( α . . . α j − ) − α − j ( j ≥ ,π j = p · · · p j , π ′ j = p ′ · · · p ′ j . Let ω ( . . . ) denote Buchstab’s function. Lemma 22. Let E be a polytope, E ⊆ P j . Let j + 1 ≤ k ≤ .(i) Let S k ( E ) = X α j ∈ E X p j ≤ p j +1 ≤···≤ p k − π k − p k − ≤ X π k − . Then S k ( E ) ≪ . OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 45 (ii) Let S ∗ k ( E ) = X α j ∈ E X p j ≤ p j +1 ≤···≤ p k − X <π k − p k − ≤ X π k − . Then S ∗ k ( E ) ≪ L − . Proof. Mertens’ formula [7, Chapter 7] implies X A
Let E be a polytope, E ⊆ P j . Let f ( E ; X ) = X α j ∈ E X j +1 ≤ k ≤ X p j ≤ p j +1 ≤···≤ p k − π k − p k − ≤ X π k − log( X/π k − ) . As X → ∞ , we have f ( E ; X ) = (1 + o (1)) 1 L Z E f j ( z j ) ω (cid:18) − z − · · · − z j z j (cid:19) dz . . . dz j . Proof. This is a slight variant of [5, Lemma 20]. (cid:3) We now discuss the major arc for Theorems 1–6, and complete theproofs of the theorems.(i) Theorem 3 . We easily verify that for functions f j (1 ≤ j ≤ s, ≤ s ≤ 5) and g havingsup x ∈ [ − τ,τ ] | f j | ≪ X, Z τ − τ | f j | dx ≪ X − c L , sup x ∈ [ − τ,τ ] | f ( x ) − g ( x ) | ≪ X exp( − C L / ) , OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 47 we have Z τ − τ g ( x ) f ( x ) . . . f s ( x )Φ( x ) e ( − Rx ) dx (9.1) − Z τ − τ f ( x ) f ( x ) . . . f s ( x )Φ( x ) e ( − Rx ) dx ≪ X s − c − cη exp( − C L / ) . Thus in view of Lemma 23 (i), we can replace R τ − τ S ( x ) Φ( x ) dx by R τ − τ I ( x ) Φ( x ) e ( − Rx ) dx , replacing factors one at a time, with error ≪ X − c − cη exp( − C L / ). Now we extend the integral to R with totalerror ≪ X − c − cη exp( − C L / ) using Lemma 19 (i) and the case s = 4of Z ∞ τ | x | − s X s − cs | Φ( x ) | dx ≪ ( X − c +8 η ) − s +1 X s − cs − cη (9.2) = X s − c − cη − s − η . We find using (1.4) and the bound (4.1) that(9.3) L A ( R ) ≫ Z ∞−∞ I ( x ) Φ( x ) e ( − Rx ) dx + O ( X − c − cη exp( − C L / )) . The integral here is Z XX Z XX Z XX Z XX φ ( t c + · · · + t c − R ) dt . . . dt (9.4) ≫ X − c − cη by Lemma 21. This yields Theorem 3 at once.(ii) Theorem 1 . If f , f , g satisfy f j ≪ X , R τ − τ f j ≪ X − c L ,sup x ∈ [ − τ,τ ] | f ( x ) − g ( x ) | ≪ X exp( − C L / ) , then the integral Z τ − τ ( f ( x ) − g ( x )) f ( x )Φ( x ) e ( − Rx ) dx is of the form b E ( R ) where E ( y ) = ( ( f ( y ) − g ( y )) f ( y )Φ( y ) ( y ∈ [ − τ, τ ])0 otherwise.By Parseval’s formula Z VV | b E ( R ) | dR < Z R | E ( y ) | dy = Z τ − τ ( f ( y ) − g ( y )) f ( y )Φ ( y ) dy ≪ X − c − cη X exp( − C L / ) ≪ X − c − cη exp( − C L / ) . Thus in two steps we can replace R τ − τ S ( x ) Φ( x ) e ( − Rx ) dx by R τ − τ I ( x ) Φ( x ) e ( − Rx ) dx with an error that is acceptable for Theorem 1. (Com-pare the discussion of E ( x ) in Section 4.) Similarly in replacing Z τ − τ I ( x ) Φ( x ) e ( − Rx ) dx by Z R I ( x ) Φ( x ) e ( − Rx ) dx we incur an error E ( x ) with Z VV | b E ( R ) | dR < Z ∞ τ | I ( y ) | | Φ( y ) | dy which from (9.2) is ≪ X − c − cη − η . Now we easily adapt the argumentleading to (9.3) to obtain L A ( R ) ≫ X − c − cη except for a set of R in [ V, V ] whose measure is ≪ V exp( − C (log V ) / ),proving Theorem 1.(iii) Theorem 5 . In the minor arc for Theorem 5, T ( x ) − S ( x ) = O ( Xτ ). We may replace R τ − τ T ( x ) Φ( x ) e ( − Rx ) dx by R τ − τ S ( x ) Φ( x ) e ( − Rx ) dx with error O ( X − c − η ) since(9.5) T − S ≪ X Xτ , Z τ − τ | T − S | dx ≪ X − c +16 η . OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 49 Now we replace R τ − τ S ( x ) Φ( x ) e ( − Rx ) dx by R τ − τ J ( x ) Φ( x ) e ( − Rx ) dx with error O ( X − c − cη exp( − C L / )) using Lemmas 18 and 23 (i). Wethen extend the integral to (cid:2) − , (cid:3) using Lemma 19 (ii); here we notethat(9.6) Z / τ x − /c dx < ( X − c +8 η ) − c +1 < X − c − η . Now we can complete the proof of Theorem 3 by drawing on Lemma20 together with (1.5) and the result of Section 6.(iv) Theorem 2 . We consider the sum S + on the major arc. Wedecompose S + into S plus O (1) sums of the type(9.7) U k ( E, x ) := X α j ∈ E X p j ≤ p j +1 ≤···≤ p kX
8, inducing an error that is O ( L − H ) byLemma 22 (ii). This produces the quantity H L X p ,...,p k − π k − log Xπ k − ! , X p ′ ,...,p ′ k − π ′ ℓ − log X/π ′ ℓ − ! , which can be calculated to within a factor 1 + o (1) using Lemma 24.Following a similar argument with the integrals in (9.9), we arrive atrepresentations, to within a factor 1 + o (1), of Z τ − τ S ( x ) S + ( x )Φ( x ) e ( − Rx ) dx, Z τ − τ S ( x ) S + ( x ) Φ( x ) e ( − Rx ) dx OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 51 of the respective forms u + H L , ( u + ) H L , where (recalling (5.15)), u + = 1 + d + d + d + d , d = Z / / dxx (1 − x ) , d = Z / / dxx (1 − x ) ,d = Z / / Z min ( x, − x ) max ( , ( − x − y )) Z y max ( − x − y, − y ) 1 xyz ω (cid:18) − x − y − zz (cid:19) dzdydx,d = Z / / Z / − xx dydxxy (1 − x − y ) . Taking into account (1.6) and the result of Section 5, we find that(9.11) A ( R ) ≫ (1 + o (1))(2 u + − ( u + ) ) H L . Using a computer calculation for d and d , we find that d < . , d < . , d < . , d < . . Thus u + ∈ (1 , Theorem 4 . The discussion of the major arc is similar to thatfor Theorem 2. We decompose S + ( x ) as S ( x ) plus O (1) sums of theform U k ( E, x ). We replace U k ( E, x ) by V k ( E, x ) with error O ( X L − ).By a variant of the argument leading to (9.1), we can replace Z τ − τ S ( x ) S + ( x )Φ( x ) e ( − Rx ) dx, Z τ − τ S ( x ) S + ( x ) Φ( x ) e ( − Rx ) dx respectively by(9.12) X ( k,E ) Z τ − τ I ( x ) U k ( E, x )Φ( x ) e ( − Rx ) dx and (9.13) X ( k,E ) X ( ℓ,E ′ ) Z τ − τ I ( x ) U k ( E, x ) U ℓ ( E ′ , x )Φ( x ) e ( − Rx ) dx with error O ( X − c − cη L − ). We extend the integrals in (9.12), (9.13) to R with error O ( X − c − cη L − ) using Lemma 19 (ii).Omitting regions of summation, we have Z R I ( x ) U k ( E, x ) U ℓ ( E ′ , x )Φ( x ) e ( − Rx ) dx = X p ,...,p k − X p ′ ,...,p ′ ℓ − π k − π ′ ℓ − L (log X ) /π k − )(log X/π ′ ℓ − ) Z XX Z XX Z X X Z X X Z X X Z ∞−∞ e ( x ( t c + t c + t c + t c + t c − R )Φ( x ) dt . . . dt where X = max (cid:0) π k − p k − , X (cid:1) , X = max (cid:0) π ′ k − p ′ k − , X (cid:1) . We writethe inner integral as φ ( t c + · · · + t c − R ) and replace X , X by X/ O ( L − H ), by Lemma 22 (ii). This producesthe quantity H L X p ,...,p k − π k − log Xπ k − ! X p ′ ,...,p ′ ℓ − π ′ ℓ − log Xπ ′ ℓ − ! . Arguing as in the preceding proof, we arrive at representations of Z τ − τ S ( x ) S + ( x )Φ( x ) e ( − Rx ) dx , Z τ − τ S ( x ) S + ( x ) Φ( x ) e ( − Rx ) dx to within a factor 1 + o (1), of the respective forms u + H L , ( u + ) H L . Here, recalling (7.6), u + = 1 + d + d ;the integrals d = Z . . dxx (1 − x ) and d = Z / . Z (1 − x ) x dy dxxy (1 − x − y ) OME DIOPHANTINE EQUATIONS AND INEQUALITIES WITH PRIMES 53 take account of the products p p ∼ X ( p ≤ p ) and p p p ∼ X ( p ≤ p ≤ p ) respectively. Simple estimations yield1 < u + < . , and Theorem 5 follows from (1.6) combined with the minor arc boundof Section 7.(vi) Theorem 6 . As in the discussion of the major arc for Theorem5, we replace T ( x ), T + ( x ) respectively by S ( x ), S + ( x ) with acceptableerror. We decompose S + as S plus two sums of the form U k ( E, x ) in(9.7), k = 2 , 3. Using Lemma 22 (ii), we replace V k ( E, x ) by W k ( E, x ) = X α ∈ E X p ≤ p ≤···≤ p k − π k − log Xπ k − ( v ( X c , x )) − v (cid:18) max (cid:18) π ck − p ck − , (cid:18) X (cid:19) c (cid:19)(cid:19) with error O ( X L − ); similarly for S ( x ). By a variant of the argumentleading to (9.1), we replace Z τ − τ S ( x ) S + ( x ) e ( − rx ) dx , Z τ − τ S ( x ) S + ( x ) e ( − rx ) dx respectively by X ( k,E ) L Z τ − τ J ( x ) W k ( E, x ) e ( − rx ) dx, (9.14) X ( k,E ) X ( ℓ,E ′ ) L Z ττ J ( x ) W k ( E, x ) W ℓ ( E ′ , x ) e ( − rx ) dx (9.15)with error O ( X − c − cη L − ). Using Lemma 19 (ii), with the same errorwe can extend the integrals in (9.14), (9.15) to (cid:2) − , (cid:3) . Thus theexpression in (9.15) has been replaced by1 L X p ,...,p k X p ′ ,...,p ′ ℓ − π k − π ′ ℓ − (cid:16) log Xπ k − (cid:17) (cid:16) log Xπ ′ ℓ − (cid:17)Z − J ( x )( v ( X c , x ) − v ( X c , x ))( v ( X c , x ) − v ( X c , x )) e ( − rx ) dx, where X = max (cid:0) π k − π k − , X (cid:1) X = max (cid:0) π ′ ℓ − p ′ ℓ − , X (cid:1) . We replace X , X by X , incurring an error that is O ( L − L ). This produces thequantity L L X p ,...,p k − π k − (cid:16) log Xπ k − (cid:17) X p ′ ,...,p ′ ℓ − π ′ ℓ − (cid:16) log Xπ ′ ℓ − (cid:17) . 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