In this paper, we mainly prove two conjectural supercongruences of Sun by using the following identity:

$$\begin{aligned} \sum _{k=0}^n\left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) ^2=16^n\sum _{k=0}^n \frac{\left( {\begin{array}{c}n+k\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2}{(-16)^k}, \end{aligned}$$

which arises from a \({}_4F_3\) hypergeometric transformation. For any prime \(p>3\), we prove that

$$\begin{aligned}&\sum _{n=0}^{p-1}\frac{n+1}{8^n}\sum _{k=0}^n\left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2 \left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) ^2\equiv (-1)^{(p-1)/2}p+5p^3E_{p-3}\pmod {p^4},\\&\sum _{n=0}^{p-1}\frac{2n+1}{(-16)^n} \sum _{k=0}^n\left( {\begin{array}{c}2k\\ k\end{array}}\right) ^2\left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) ^2\equiv (-1)^{(p-1)/2}p +3p^3E_{p-3}\pmod {p^4}, \end{aligned}$$

where \(E_{p-3}\) is the \((p-3)\hbox {th}\) Euler number.

中文翻译：

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关于Z.-W.的两个猜想超同余。太阳

在本文中，我们主要通过使用以下恒等式证明Sun的两个猜想超同余：

$$ \ begin {aligned} \ sum _ {k = 0} ^ n \ left（{\ begin {array} {c} 2k \\ k \ end {array}} \ right）^ 2 \ left（{\ begin {array} {c} 2n-2k \\ nk \ end {array}} \ right）^ 2 = 16 ^ n \ sum _ {k = 0} ^ n \ frac {\ left（{\ begin {array} { c} n + k \\ k \ end {array}} \ right）\ left（{\ begin {array} {c} n \\ k \ end {array}} \ right）\ left（{\ begin {array } {c} 2k \\ k \ end {array}} \ right）^ 2} {（-16）^ k}，\ end {aligned} $$

它来自\（{} _ 4F_3 \）超几何变换。对于任何素数\（p> 3 \），我们证明

$$ \ begin {aligned}＆\ sum _ {n = 0} ^ {p-1} \ frac {n + 1} {8 ^ n} \ sum _ {k = 0} ^ n \ left（{\ begin {array} {c} 2k \\ k \ end {array}} \ right）^ 2 \ left（{\ begin {array} {c} 2n-2k \\ nk \ end {array}} \ right）^ 2 \ equiv（-1）^ {（p-1）/ 2} p + 5p ^ 3E_ {p-3} \ pmod {p ^ 4}，\\＆\ sum _ {n = 0} ^ {p-1 } \ frac {2n + 1} {（-16）^ n} \ sum _ {k = 0} ^ n \ left（{\ begin {array} {c} 2k \\ k \ end {array}} \ right ）^ 2 \ left（{\ begin {array} {c} 2n-2k \\ nk \ end {array}} \ right）^ 2 \ equiv（-1）^ {（p-1）/ 2} p + 3p ^ 3E_ {p-3} \ pmod {p ^ 4}，\ end {aligned} $$

其中\（E_ {p-3} \）是\（（p-3）\ hbox {th} \）欧拉数。