<p>Employing Gasper and Rahman’s quadratic summation, the method of “creative microscoping” developed by the first author and Zudilin, and the Chinese remainder theorem for coprime polynomials, we prove some new <i>q</i>-supercongruences modulo the third and fourth powers of a cyclotomic polynomial. As a conclusion, we obtain some new supercongruences, such as: for primes <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="25_2025_2482_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="50" /> </InlineMediaObject> <EquationSource Format="TEX">$p\geqslant 13$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>⩾</mo> <mn>13</mn> </mrow> </math></EquationSource> </InlineEquation> with <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="25_2025_2482_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="106" /> </InlineMediaObject> <EquationSource Format="TEX">$p\equiv 1\pmod 4$</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>4</mn> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>, <Equation ID="Equ33"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="25_2025_2482_Article_Equ33.gif" Format="GIF" Height="54" Rendition="HTML" Resolution="72" Type="Linedraw" Width="304" /> </MediaObject> <EquationSource Format="TEX">$\begin{aligned} \sum _{k=0}^{(3p+1)/4}(6k+1)\frac{(\frac{1}{2})_k^3(-\frac{1}{4})_k}{(k+1)!k!^3 4^k}\equiv 0\pmod {p^4}, \end{aligned}$</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> <mo stretchy="false">(</mo> <mn>3</mn> <mi>p</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mn>4</mn> </mrow> </munderover> <mrow> <mo stretchy="false">(</mo> <mn>6</mn> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mfrac> <mrow> <msubsup> <mrow> <mo stretchy="false">(</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mi>k</mi> <mn>3</mn> </msubsup> <msub> <mrow> <mo stretchy="false">(</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> <mi>k</mi> </msub> </mrow> <mrow> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mo>!</mo> <mi>k</mi> <msup> <mo>!</mo> <mn>3</mn> </msup> <msup> <mn>4</mn> <mi>k</mi> </msup> </mrow> </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> <mn>4</mn> </msup> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>where <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="25_2025_2482_Article_IEq3.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="59" /> </InlineMediaObject> <EquationSource Format="TEX">$(a)_0=1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo stretchy="false">)</mo> </mrow> <mn>0</mn> </msub> <mo>=</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="25_2025_2482_Article_IEq4.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="215" /> </InlineMediaObject> <EquationSource Format="TEX">$(a)_k = a(a+1)\cdots (a+k-1)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo stretchy="false">)</mo> </mrow> <mi>k</mi> </msub> <mo>=</mo> <mi>a</mi> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mo>⋯</mo> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo>+</mo> <mi>k</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> for <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="25_2025_2482_Article_IEq5.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">$k\geqslant 1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>⩾</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>.</p>

Some New q-supercongruences Arising from a Quadratic Summation of Gasper and Rahman

期刊论文
发表日期：

2025年7月26日

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Victor J. W. Guo,
Xin Zhao

摘要

Employing Gasper and Rahman’s quadratic summation, the method of “creative microscoping” developed by the first author and Zudilin, and the Chinese remainder theorem for coprime polynomials, we prove some new q-supercongruences modulo the third and fourth powers of a cyclotomic polynomial. As a conclusion, we obtain some new supercongruences, such as: for primes $p\geqslant 13$ $p ⩾ 13$ with $p\equiv 1\pmod 4$ $p \equiv 1 (mod 4)$ , $\begin{aligned} \sum _{k=0}^{(3p+1)/4}(6k+1)\frac{(\frac{1}{2})_k^3(-\frac{1}{4})_k}{(k+1)!k!^3 4^k}\equiv 0\pmod {p^4}, \end{aligned}$ $\begin{matrix} \sum_{k = 0}^{(3 p + 1) / 4} (6 k + 1) \frac{{(\frac{1}{2})}_{k}^{3} {(- \frac{1}{4})}_{k}}{(k + 1)! k!^{3} 4^{k}} \equiv 0 (mod p^{4}), \end{matrix}$ where $(a)_0=1$ ${(a)}_{0} = 1$ and $(a)_k = a(a+1)\cdots (a+k-1)$ ${(a)}_{k} = a (a + 1) \dots (a + k - 1)$ for $k\geqslant 1$ $k ⩾ 1$ .