In this paper, by applying the q-Dixon formula, the ‘creative microscoping’ method introduced by Guo and Zudilin and the Chinese remainder theorem for coprime polynomials, we prove Wei’s conjecture: for any positive integer s and any prime p with <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="40840_2025_1919_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="106" /> </InlineMediaObject> <EquationSource Format="TEX">$p \equiv 1 \pmod 6$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>≡</mo> <mn>1</mn> <mspace width="4.44443pt" /> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <mn>6</mn> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>, <Equation ID="Equ21"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="40840_2025_1919_Article_Equ21.gif" Format="GIF" Height="54" Rendition="HTML" Resolution="72" Type="Linedraw" Width="210" /> </MediaObject> <EquationSource Format="TEX">$\begin{aligned} \sum _{k=0}^{\left( p^s+2\right) / 3} \frac{\left( -\frac{2}{3}\right) _k^3}{(1)_k^3} \equiv 0 \pmod { p^{3 s}}. \end{aligned}$</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mfenced close=")" open="("> <msup> <mi>p</mi> <mi>s</mi> </msup> <mo>+</mo> <mn>2</mn> </mfenced> <mo stretchy="false">/</mo> <mn>3</mn> </mrow> </munderover> <mfrac> <msubsup> <mfenced close=")" open="("> <mo>-</mo> <mfrac> <mn>2</mn> <mn>3</mn> </mfrac> </mfenced> <mi>k</mi> <mn>3</mn> </msubsup> <msubsup> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mi>k</mi> <mn>3</mn> </msubsup> </mfrac> <mo>≡</mo> <mn>0</mn> <mspace width="10.0pt" /> <mrow> <mo stretchy="false">(</mo> <mo>mod</mo> <mspace width="0.277778em" /> <msup> <mi>p</mi> <mrow> <mn>3</mn> <mi>s</mi> </mrow> </msup> <mo stretchy="false">)</mo> </mrow> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>

Proof of a conjecture by Wei

期刊论文
发表日期：

2025年6月26日

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Yu Hu,
Xiaoxia Wang

摘要

In this paper, by applying the q-Dixon formula, the ‘creative microscoping’ method introduced by Guo and Zudilin and the Chinese remainder theorem for coprime polynomials, we prove Wei’s conjecture: for any positive integer s and any prime p with $p \equiv 1 \pmod 6$ $p \equiv 1 (mod 6)$ , $\begin{aligned} \sum _{k=0}^{\left( p^s+2\right) / 3} \frac{\left( -\frac{2}{3}\right) _k^3}{(1)_k^3} \equiv 0 \pmod { p^{3 s}}. \end{aligned}$ $\begin{matrix} \sum_{k = 0}^{(p^{s} + 2) / 3} \frac{{(- \frac{2}{3})}_{k}^{3}}{{(1)}_{k}^{3}} \equiv 0 (mod p^{3 s}) . \end{matrix}$