On divisibility of sums of Apery polynomials

For any positive integers $m$ and $α$, we prove that $\sum_{k=0}^{n-1}ε^k(2k+1)A_k^{(α)}(x)^m\equiv0\pmod{n},$ where $ε\in\{1,-1\}$ and $A_n^{(α)}(x)=\sum_{k=0}^n\binom{n}{k}^α\binom{n+k}{k}^αx^k.$

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