This is Voronoi’s congruence (see e.g. [ 2 , Chapter 15 ] ). Writing \(ja = p\lfloor ja/p \rfloor + r_j\) with \(r_j\) the residue of \(ja\), summing \(r_j^k\) over \(j\) and expanding by the binomial theorem identifies the two sides modulo \(p\), using Faulhaber’s formula for power sums and the von Staudt–Clausen integrality (lemma 2.5) to control the Bernoulli denominators of the lower-order terms.

Choose a primitive root \(a\) modulo \(p\) and apply theorem 5.1 to both indices. Since \(m \equiv n \pmod{p-1}\), Fermat’s little theorem gives \(a^m \equiv a^n\) and \(j^{m-1}\equiv j^{n-1} \pmod p\) for all \(j\), so the two Voronoi sums agree modulo \(p\). Subtracting the two instances yields \((a^{n}-1)\bigl(B_m/m - B_n/n\bigr) \equiv 0 \pmod p\), and because \(a\) is a primitive root with \((p-1)\nmid n\) the factor \(a^{n}-1\) is a \(p\)-adic unit, giving the congruence.

It remains to lift the restriction \(p \ge 5\) of theorem 5.2 to all odd primes. The only other odd prime is \(p = 3\), where \(p - 1 = 2\) divides every even index; the hypothesis \((p-1)\nmid n\) is then unsatisfiable, so there is nothing left to prove.