In this paper, we mainly prove two conjectural supercongruences of Sun by using the following identity: ∑k=0n2kk22n-2kn-k2=16n∑k=0nn+kknk2kk2(-16)k,\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \sum _{k=0}^n\left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) ^2=16^n\sum _{k=0}^n \frac{\left( {\begin{array}{c}n+k\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2}{(-16)^k}, \end{aligned}$$\end{document}which arises from a 4F3\documentclass[12pt]{minimal}
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\begin{document}$${}_4F_3$$\end{document} hypergeometric transformation. For any prime p>3\documentclass[12pt]{minimal}
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\begin{document}$$p>3$$\end{document}, we prove that ∑n=0p-1n+18n∑k=0n2kk22n-2kn-k2≡(-1)(p-1)/2p+5p3Ep-3(modp4),∑n=0p-12n+1(-16)n∑k=0n2kk22n-2kn-k2≡(-1)(p-1)/2p+3p3Ep-3(modp4),\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned}&\sum _{n=0}^{p-1}\frac{n+1}{8^n}\sum _{k=0}^n\left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2 \left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) ^2\equiv (-1)^{(p-1)/2}p+5p^3E_{p-3}\pmod {p^4},\\&\sum _{n=0}^{p-1}\frac{2n+1}{(-16)^n} \sum _{k=0}^n\left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) ^2\equiv (-1)^{(p-1)/2}p +3p^3E_{p-3}\pmod {p^4}, \end{aligned}$$\end{document}where Ep-3\documentclass[12pt]{minimal}
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\begin{document}$$E_{p-3}$$\end{document} is the (p-3)th\documentclass[12pt]{minimal}
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\begin{document}$$(p-3)\hbox {th}$$\end{document} Euler number.
