Some <i>q</i> -supercongruences from a~very-well-poised ₆ φ ₅ summation

Abstract By making use of a very-well-poised <m:math xmlns:m="http://www.w3.org/1998/Math/MathML"> <m:mmultiscripts> <m:mi>ϕ</m:mi> <m:mn>5</m:mn> <m:none/> <m:mprescripts/> <m:mn>6</m:mn> <m:none/> </m:mmultiscripts> </m:math> {{}_{6}\phi_{5}} summation, the creative microscoping method, and the Chinese remainder theorem for coprime polynomials, we establish some new q -supercongruences, including some Dwork-type q -supercongruences. As a consequence, we obtain the following result: for any prime <m:math xmlns:m="http://www.w3.org/1998/Math/MathML"> <m:mrow> <m:mi>p</m:mi> <m:mo>≡</m:mo> <m:mrow> <m:mn>1</m:mn> <m:mspace width="veryverythickmathspace"/> <m:mrow> <m:mo lspace="8.1pt" stretchy="false">(</m:mo> <m:mrow> <m:mo>mod</m:mo> <m:mn>4</m:mn> </m:mrow> <m:mo stretchy="false">)</m:mo> </m:mrow> </m:mrow> </m:mrow> </m:math> {p\equiv 1\pmod{4}} and integer <m:math xmlns:m="http://www.w3.org/1998/Math/MathML"> <m:mrow> <m:mi>r</m:mi> <m:mo>⩾</m:mo> <m:mn>1</m:mn> </m:mrow> </m:math> {r\geqslant 1} , <m:math xmlns:m="http://www.w3.org/1998/Math/MathML"> <m:mrow> <m:mrow> <m:mrow> <m:munderover> <m:mo largeop="true" movablelimits="false" symmetric="true">∑</m:mo> <m:mrow> <m:mi>k</m:mi> <m:mo>=</m:mo> <m:mn>0</m:mn> </m:mrow> <m:mfrac> <m:mrow> <m:msup> <m:mi>p</m:mi> <m:mi>r</m:mi> </m:msup> <m:mo>-</m:mo> <m:mn>1</m:mn> </m:mrow> <m:mn>2</m:mn> </m:mfrac> </m:munderover> <m:mrow> <m:mrow> <m:mo stretchy="false">(</m:mo> <m:mrow> <m:mrow> <m:mn>8</m:mn> <m:mo>⁢</m:mo> <m:mi>k</m:mi> </m:mrow> <m:mo>+</m:mo> <m:mn>1</m:mn> </m:mrow> <m:mo stretchy="false">)</m:mo> </m:mrow> <m:mo>⁢</m:mo> <m:mfrac> <m:mrow> <m:msubsup> <m:mrow> <m:mo stretchy="false">(</m:mo> <m:mfrac> <m:mn>1</m:mn> <m:mn>4</m:mn> </m:mfrac> <m:mo stretchy="false">)</m:mo> </m:mrow> <m:mi>k</m:mi> <m:mn>3</m:mn> </m:msubsup> <m:mo>⁢</m:mo> <m:msub> <m:mrow> <m:mo stretchy="false">(</m:mo> <m:mfrac> <m:mn>1</m:mn> <m:mn>2</m:mn> </m:mfrac> <m:mo stretchy="false">)</m:mo> </m:mrow> <m:mi>k</m:mi> </m:msub> </m:mrow> <m:mrow> <m:msup> <m:mrow> <m:mi>k</m:mi> <m:mo lspace="0pt" rspace="3.5pt">!</m:mo> </m:mrow> <m:mn>3</m:mn> </m:msup> <m:mo>⁢</m:mo> <m:msub> <m:mrow> <m:mo stretchy="false">(</m:mo> <m:mfrac> <m:mn>3</m:mn> <m:mn>4</m:mn> </m:mfrac> <m:mo stretchy="false">)</m:mo> </m:mrow> <m:mi>k</m:mi> </m:msub> </m:mrow> </m:mfrac> </m:mrow> </m:mrow> <m:mo>≡</m:mo> <m:mrow> <m:msup> <m:mi>p</m:mi> <m:mi>r</m:mi> </m:msup> <m:mspace width="veryverythickmathspace"/> <m:mrow> <m:mo lspace="8.1pt" stretchy="false">(</m:mo> <m:mrow> <m:mo movablelimits="false">mod</m:mo> <m:msup> <m:mi>p</m:mi> <m:mrow> <m:mi>r</m:mi> <m:mo>+</m:mo> <m:mn>3</m:mn> </m:mrow> </m:msup> </m:mrow> <m:mo stretchy="false">)</m:mo> </m:mrow> </m:mrow> </m:mrow> <m:mo>,</m:mo> </m:mrow> </m:math> \sum_{k=0}^{\frac{p^{r}-1}{2}}(8k+1)\frac{(\frac{1}{4})_{k}^{3}(\frac{1}{2})_{% k}}{k!^{3}(\frac{3}{4})_{k}}\equiv p^{r}\pmod{p^{r+3}},