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