A q-congruence implying the Beukers–Van Hamme congruence

By making use of Andrews' terminating q-analogue of Watson's formula and a double sum identity, we give a q-analogue of the following congruence: for any prime $p\equiv 1\left(\mathrm{mod}4\right)$,$\sum _{k=0}^{(p-1)/2}\left(\begin{array}{c}(p-1)/2\\ k\end{array}\right)\left(\begin{array}{c}(p-1)/2+k\\ k\end{array}\right)\equiv \frac{1}{2^{(p-1)/2}}\left(\begin{array}{c}(p-1)/2\\ (p-1)/4\end{array}\right)\left(\mathrm{mod}{p}^2\right).$ In view of the Chowla–Dwork–Evans congruence, our q-congruence may somewhat be regarded as a q-analogue of the Beukers–Van Hamme congruence:$\sum _{k=0}^{(p-1)/2}\left(\begin{array}{c}(p-1)/2\\ k\end{array}\right)\left(\begin{array}{c}(p-1)/2+k\\ k\end{array}\right)\equiv {(-1)}^{(p-1)/4}(2a-\frac{p}{2a})\left(\mathrm{mod}{p}^2\right),$ where $p=a^2+b^2$ with $a,b\in Z$ and $a\equiv 1\left(\mathrm{mod}4\right)$.