Note on super congruences modulo $p^2$

Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak\binom{-1-a}k}{m^k} \equiv 0\pmod p \quad\hbox{implies}\quad\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak \binom{-1-a}k}{m^k}\equiv 0\pmod {p^2}.$

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