  Let $p$ be an odd prime and let $n$ be a positive integer. For any positive
integer $\alpha$ and $m\in\{1,2,3\}$, we have \begin{align*}
\sum_{k=0}^{p^{\alpha}n-1}\frac{(\frac12)_k}{k!}\cdot\frac{(-4)^k}{m^k}\equiv\bigg(\frac{m(m-4)}{p}\bigg)\sum_{k=0}^{p^{\alpha-1}n-1}\frac{(\frac12)_k}{k!}\cdot\frac{(-4)^k}{m^k}\pmod{p^{2\alpha}},
\end{align*} where $(x)_k=x(x+1)\cdots(x+k-1)$ and
$\big(\frac{\cdot}{\cdot}\big)$ denotes the Legendre symbol. Also, when $m=4$,
\begin{align*}
\sum_{k=0}^{p^{\alpha}n-1}(-1)^k\cdot\frac{(\frac12)_k}{k!}\equiv
p\sum_{k=0}^{p^{\alpha-1}n-1}(-1)^k\cdot\frac{(\frac12)_k}{k!}\pmod{p^{2\alpha}}.
\end{align*}
