Motivated by the recent work of He and Wang and of Guo and Zhu on q-congruences, we present two new q-congruences on double basic hypergeometric sums. As a conclusion, we obtain the following result: for odd primes \(p\equiv 1\pmod {4}\) , \( \sum _{k=0}^{(p-1)/2}(6k+1)\frac{(\frac{1}{2})_k(\frac{1}{4})_k^2}{k!^3}4^k \sum _{j=1}^{k}\left\{ \frac{1}{(4j-3)^2}-\frac{1}{4j^2}\right\} \equiv 0\pmod {p^2}, \) where \((a)_k=a(a+1)\cdots (a+k-1)\) . This result is similar to a congruence conjectured by Long and later confirmed by Swisher.
