Even-modulus power-sum binomial congruence

About 22 years old · traced to 2004

Let $S_m(n)=\sum_{j=1}^{n-1}j^m$ , and let $k$ be a positive integer. Even-modulus power-sum conjecture. For every $k\ge 1$ and every even positive integer $n$ , the congruence

\sum_{i=0}^k\left(1+n(k+1-i)\right)\binom{2k+2}{2i}S_{2i}(n)\equiv -1\pmod{n^3}

holds. The source notes that division by $2$ is immediate for odd $n$ and proposes the even- $n$ extension based on computations for small values of $k$ ; no proof is given in the supplied text.

References

Primary source

Romeo Meštrović, “A congruence modulo n^3 involving two consecutive sums of powers and its applications”, arXiv:1211.4570 (2012).

Additional references

2 papers in this index state this conjecture (2004–2012). The statement above is taken from the most recent of them; the others are arXiv:hep-th/0404194.

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader has posted a complete proof of the even case, but it has not been independently checked, so the conjecture is not yet settled.

Romeo Meštrović proposed this as Conjecture 2.17 in 2012: for every $k\ge 1$ and even positive integer $n$ , the stated power-sum expression is congruent to $-1$ modulo $n^3$ . The paper gave computational evidence but no proof.

Known results

Meštrović, 2012: the corresponding congruence after multiplying by $2$ was proved modulo $n^3$ .
Meštrović, 2012: division by $2$ yields the conjectured congruence for odd $n$ .
The even case was supported only by computations for small $k$ and even $n$ .

Posted attempt

A reader claims a complete proof for even $n$ , reducing the assertion to divisibility of a weighted sum by $2n^3$ , then treating paired terms $x$ and $n-x$ and the central term. The argument has not been independently verified.

Current status (as of August 2026): The odd- $n$ case and the doubled congruence are established, while a complete even- $n$ proof has been posted but remains unverified.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

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.