Even-modulus power-sum binomial congruence

Write S_r(n)=Σ_{x=1}^{n−1}x^r, including S_0(n)=n−1. Fix k≥1 and an even positive n, put m=2k+2, and set A=Σ_{i=0}^k [1+n(k+1−i)]C(m,2i)S_{2i}(n). We prove A≡−1 mod n³.

The identity (2i+1)C(m,2i+1)=2(m/2−i)C(m,2i) and Pascal telescoping Σ_{r=0}^{m−1}C(m,r)S_r(n)=n^m−1 give the exact equality A+1=n^m−W/2, W=Σ_{i=0}^k C(m,2i+1)D_i, D_i=2S_{2i+1}(n)−(2i+1)nS_{2i}(n). Since m≥4, it suffices to prove 2n³|W. Note that D_0=0.

Write n=2c and f_i(x)=2x^{2i+1}−(2i+1)nx^{2i}. For i≥1, expansion at the paired arguments x and n−x gives f_i(x)+f_i(n−x)≡−n³C(2i+1,3)x^{2i−2} mod 2n³; all lower powers of n cancel, and all remaining terms contain n⁴, which is divisible by 2n³ because n is even. Therefore the weighted contribution of each pair 1≤x<c is −n³Σ_{i=1}^k C(m,2i+1)C(2i+1,3)x^{2i−2} =−n³C(m,3)Σ_{i=1}^k C(m−3,2i−2)x^{2i−2} modulo 2n³. Since m is even, C(m,3) is even, so every paired contribution vanishes modulo 2n³.

For the unpaired central argument x=c, f_i(c)=−4ic^{2i+1}. We claim 4|iC(m,2i+1) for every even m and i≥1. The binomial coefficient is even because its upper index is even and its lower index odd, proving the claim when i is even. If i is odd, use (2i+1)C(m,2i+1)=mC(m−1,2i). For 4|m the claim is immediate. Otherwise m≡2 mod4; then m−1 has binary 2's digit 0 while 2i has binary 2's digit 1, so Lucas's theorem gives C(m−1,2i) even, again proving the claim. Hence 2n³=16c³ divides C(m,2i+1)f_i(c) for every i≥1.

All paired and central contributions are therefore divisible by 2n³, proving 2n³|W. Thus n³|A+1, as required for every k≥1 and every even n. Together with the already proved odd-modulus case, the same congruence holds for all positive n.