On divisibility of sums of Apéry polynomials

For any positive integers m and α, we prove that∑k=0n−1ϵk(2k+1)Ak(α)(x)m≡0(modn), where ϵ∈{1,−1} and the generalized Apéry polynomialAn(α)(x)=∑k=0n(nk)α(n+kk)αxk. The key to our proof is to use q-congruences.