The Waring-Goldbach Problem In Short Intervals

Suppose that k≥ 2 is a positive integer and p is a prime.Let τ=τ(k,p)be the integer such that pτ‖ k,which means that pτ|k but pτ+1(?)k.Define γ=γ((k,p)by taking#12 We now put Rk=(?)pγ.(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<θ<1,we wonder for every sufficiently large n≡s(mod Rk)if there exists primes p1,…,ps∈[x-xθ,x+xθ]such that p1k+…+psk=n.Let us define the parameter 0<θk,s<1 by the following way:the equation#12 is solvable for all θ>θk,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{θ,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 ε>0 andθ∈(1/2,1).Let α-be a positive number such that for any large enough x,any interval I(?)[x,x+xθ+∈]with |I|≥ xθ-ε,and any c,d∈N with(c,d)=1 and d ≤ log x we always have(?)1≥α-|I|/Φ(d)logx Then when s>max{2/α-θ,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-nθ/k,(n/s)1/k+nθ/k]such that n=p1k+…psk.[13,Theorem 10.3]yields that when θ>11/20,the parameter α-in above theorem can be taken α-=99/100.On recalling our definition of Ok,s,it is immediate that Theorem 0.1 gives θk,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 θ∈(1/2,1).Let α-be the parameter defined in Theorem 0.1.Then when s>max{2/α-θ,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|≤nθ/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.Letθk,s*be the smallest θ such that the following equation is solvable for almost all n∈(M-(M/s)k-1+θ/k,M]#12Then one can deduce from Theorem 0.2 and[13,Theorem 10.3]that θk,k(k+1)/2+1*<0.55+∈when k≥ 2.This improves[20,Theorem 2]who proved that θk,k(k+1)/2+1*≤31/40+∈.