  For any $m,n\in\mathbb{N}=\{0,1,2\ldots\}$, the truncated hypergeometric
series ${}_{m+1}F_m$ is defined by $$
{}_{m+1}F_m\bigg[\begin{matrix}x_0&x_1&\ldots&x_m\\
&y_1&\ldots&y_m\end{matrix}\bigg|z\bigg]_n=\sum_{k=0}^n\frac{(x_0)_k(x_1)_k\cdots(x_m)_k}{(y_1)_k\cdots(y_m)_k}\cdot\frac{z^k}{k!},
$$ where $(x)_k=x(x+1)\cdots(x+k-1)$ is the Pochhammer symbol. Let $p$ be an
odd prime. For $\alpha,z\in\mathbb{Z}_p$ with $\langle
-\alpha\rangle_p\equiv0\pmod{2}$, where $\langle x\rangle_p$ denotes the least
nonnegative residue of $x$ modulo $p$ for any $x\in\mathbb{Z}_p$, we mainly
prove the following congruence motivated by Orr's identity: $$
{}_2F_1\bigg[\begin{matrix}\frac12\alpha&\frac32-\frac12\alpha\\
&1\end{matrix}\bigg|z\bigg]_{p-1}{}_2F_1\bigg[\begin{matrix}\frac12\alpha&\frac12-\frac12\alpha\\
&1\end{matrix}\bigg|z\bigg]_{p-1}\equiv{}_3F_2\bigg[\begin{matrix}\alpha&2-\alpha&\frac12\\
&1&1\end{matrix}\bigg|z\bigg]_{p-1}\pmod{p^2}. $$ As a corollary, for any
positive integer $b$ with $p\equiv\pm1\pmod{b}$ and $\langle
-1/b\rangle_p\equiv0\pmod{2}$, we deduce that $$
\sum_{k=0}^{p-1}(b^2k+b-1)\frac{\binom{2k}{k}}{4^k}\binom{-1/b}{k}\binom{1/b-1}{k}\equiv0\pmod{p^2}.
$$ This confirms a conjectural congruence of the second author.
