We give two q-congruences on double basic hypergeometric sums. As a conclusion, we obtain the following supercongruence: for any odd prime p, \(\begin{aligned}&\sum _{k=0}^{p-1}(6k+1)\frac{(\frac{1}{2})_k^3(\frac{1}{4})_k^2 }{k!^5 }\sum _{j=1}^{k}\left( \frac{1}{(2j-1)^2}-\frac{1}{16j^2}\right) \nonumber \\&\quad \equiv {\left\{ \begin{array}{ll} p\dfrac{(\frac{1}{2})_{(p-1)/4}^2}{(1)_{(p-1)/4}^2}\displaystyle \sum _{j=1}^{(p-1)/2}\dfrac{(-1)^{j-1}}{4j^2} \pmod {p^2}, & \text {if }p\equiv 1\pmod {4},\\ \dfrac{3(\frac{1}{2})_{(3p-1)/4}^2}{4p(1)_{(3p-1)/4}^2}\pmod {p^2}, & \text {if }p\equiv 3\pmod {4}. \end{array}\right. } \end{aligned}\) where \((a)_k=a(a+1)\cdots (a+k-1)\) . We also pose two conjectures on q-congruences for further study.
