The Waring-Goldbach Problem In Short Intervals

Suppose that k? 2 is a positive integer and p is a prime.Let ?=?(k,p)be the integer such that p?? k,which means that p?|k but p?+1(?)k.Define ?=?((k,p)by taking#12 We now put Rk=(?)p?.(0.3)Waring-Goldbach problem asks whether sufficiently large number n with n?s(mod Rk)can be expressed as a sum of s terms of k-th powers of primes,where s is a positive num-ber depending on k.The first result was obtained by Hua[14],who showed that when s?2k+1 every sufficiently large natural number n with n?s(mod Rk)can be repre-sented as n=p1k+…+psk,(0.4)where p1,···,ps are prime numbers.Subsequent works focus on reducing the value of s.For convenience,suppose that H(k)is the least integer s such that the equation(0.4)is solvable for all sufficiently large number n?s(mod Rk).The current records are as follows.When k=2,3,Hua's above result is still sharpest.Zhao[44]proved that H(4)?13 and H(6)? 32;Kawada and Wooley[17]proved that H(5)?21;Kumchev[18]showed that H(7)?46;and Kumchev and Wooley[21]proved that H(8)?61.Besides,Kumchev and Wooley[21]also showed that for the large value of k,H(k)?(4k-2)log k+k-7.One interesting generalization of Waring-Goldbach problem is to restrict the set of k-th power of primes into some sparse subsets of k-th power of prime numbers.One natural way to choose such a subset is to restrict the k-th power of primes to be all around n/s.In other word,let x=(n/s)1/k and 1/2<?<1,we wonder for every sufficiently large n?s(mod Rk)if there exists primes p1,…,ps?[x-x?,x+x?]such that p1k+…+psk=n.Let us define the parameter 0<?k,s<1 by the following way:the equation#12 is solvable for all ?>?k,s and all sufficiently large n with n? s(mod Rk).There are significant pieces of literature to study this problem.And the developments mainly come from two aspects.One is to enlarge the major arcs,the other is to improve the minor-arc estimates.Wei and Wooley[41]proved that when s>max{?,2k(k-1)},(?);This result has been improved by Huang[16],Kumchev and Liu[20]successively with the help of recent developments of Vinogradov mean value theorem,Harman's sieve methods and other skills.Afterwards,Matomaki and Shao[30]improved the bound once again when they studied the discorrelation between Mangoldt function and polynomial phases in short intervals.Matomaki and Shao showed that when k? 2 and s>k(k+1)+2 one could get Ok,s?2/3.Whilst,Matomaki,Maynard and Shao[29]used the transference principle to prove that every sufficiently large odd integer n can be represented as the sum of three prime integers which are in the interval[n/3-n0.55+?,n/3+n0.55+?].In 2019,Salmensuu[35]generalised the argument of Matomaki-Maynard-Shao to Waring-Goldbach problem in short intervals and he proved that when s>k(k+1),#12We optimize the pseudorandom condition in the transference principle,and this will lead to a shorter admissible interval.Our first result is the following.Theorem 0.1.Suppose that 2?k<s are positive integers.Suppose that ?>0 and??(1/2,1).Let ?-be a positive number such that for any large enough x,any interval I(?)[x,x+x?+?]with |I|? x?-?,and any c,d?N with(c,d)=1 and d ? log x we always have(?)1??-|I|/?(d)logx Then when s>max{2/?-?,k(k+1)}every sufficiently large integer n?s(mod Rk),where Rk is defined in(0.3),there exist primes p1,…,ps?[(n/s)1/k-n?/k,(n/s)1/k+n?/k]such that n=p1k+…psk.[13,Theorem 10.3]yields that when ?>11/20,the parameter ?-in above theorem can be taken ?-=99/100.On recalling our definition of Ok,s,it is immediate that Theorem 0.1 gives ?k,k(k+1)+1? 0.55(k? 2)whenever k? 2.This improves Salmensuu's result when k=2,3.In traditional circle method,we can usually get a result of exceptional sets by making use of Bessel's inequality(see[41,Section 9]as an example).Here we handle the excep-tional sets by establishing a quantitative almost all version of the transference principle.Theorem 0.2.Suppose that 2 ?k<s are positive integers.Suppose that ?>0 and ??(1/2,1).Let ?-be the parameter defined in Theorem 0.1.Then when s>max{2/?-?,k(k+1)/2},for almost all integers n?s(mod Rk)(besides,in the case k=3 and s=7 we also need n(?)0(mod 9)),one can find primes p1,…,ps in the interval|(n/s)1/k-pi|?n?/k(1?i?s)such that n=p1k+…+psk.We remark that the additional congruence condition n(?)0(mod 9)is to ensure p13+…+p73=n,is solvable.Now,let M be a sufficiently large positive number.Let?k,s*be the smallest ? such that the following equation is solvable for almost all n?(M-(M/s)k-1+?/k,M]#12Then one can deduce from Theorem 0.2 and[13,Theorem 10.3]that ?k,k(k+1)/2+1*<0.55+?when k? 2.This improves[20,Theorem 2]who proved that ?k,k(k+1)/2+1*?31/40+?.