YaoJie Guo | Searcharxiv

arXiv2026

The Waring-Goldbach problem in short intervals under the Generalized Riemann Hypothesis

Let N be a sufficiently large odd integer. We study the representation of a large integer m N^k as a sum of m almost equal k-th powers of primes, where the common distribution scale is at least N^theta. Assuming the Generalized Riemann Hypothesis (GRH), we prove that for any epsilon > 0 one may take theta > 1/2, provided that m < k(k+1)/2 and that m, k are odd integers sufficiently large. This is achieved by employing a new tool closely related to earlier work. For sufficiently large m, by refining the discussion of the minor arcs, we further show that under GRH the equation has a solution provided m > 16k log k + 4k + 2. Moreover, we give an asymptotic formula for the number of solutions, with main term involving N^{2k theta (1 - (1 - 1/k)^{m_1}) + theta(m-1) + 1 - k}, where m_1 = 4k. Under GRH and for sufficiently large m, k, the short-interval Waring-Goldbach problem is essentially settled.

math.NT↗

arXiv2026

An Integral Mean Value Theorem for Weyl Sums over Broken Arcs

In this article, we research the mean value of integral of exponential sum $S(α)=\sum_{u\in I}e(αu^k)$, where $I$ is a short interval whose length is $2N^θ,θ<1$, and the broken arc $\mathfrak{m^*}$ is a subset of a following minor arcs \[ \mathfrak{m}=\bigcap_{j\leq k-1}\left\{α:\forall q<(\log N)^A,h \frac{1}{qN^{(k-j-1/2)θ}}\right\} \] which has measure at least $c>0$. By setting $m$ is a sufficiently large number, $N$ is be sufficiently large in terms of $m$. When $k>\log m$ and $\frac{\log k}{\log m}<1/2$ we set the following estimate: \[ \int_{\mathfrak{m^*}}\bigg|\sum\limits_{{N_1}<u<{N_2}}{e(zu^k)}\bigg|^{m}\mathrm{d}z\ll_cN^{θm(\frac{2k+1}{2k+2}+o(1))} \] We can find this estimate moving beyond the even-odd restriction of powers. To get this bound, we first set a strong estimate for almost $α$ by Diophantine approximation and Vinogradov's main value theorem. Then, combining this result, we construct a refined Weyl differencing argument by partitioning the differences step into large and small range, which significantly outperforms the classical one. By the version of probability, we can calculate the multiplicity of each sum. Put them together and we can complete the proof.

math.NT↗