Divisibility of (finite) power sum of integers

Question

Consider the power sum $$S_a(b)=1^{2b}+2^{2b}+\cdots+(3a-2)^{2b}.$$ Let $\nu_3(x)$ denote the $3$-adic valuation of $x$.

QUESTION 1. (milder) Is this true? $$\nu_3\left(\frac{S_a(b)}{S_a(1)}\right)=0.$$ QUESTION 2. Is this true? $\nu_3(S_a(b))=\nu_3(2a-1)$.

Answer 1

Notice that $$2S_a(b) \equiv 1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \pmod{6a-3}.$$

From Faulhaber's formula, we have $$1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \equiv B_{2b} (6a-3) \pmod{6a-3}.$$ It follows that $$S_a(b) \equiv \frac32B_{2b} (2a-1)\pmod{3(2a-1)},$$ where $\nu_3(\frac32 B_{2b})=0$ by von Staudt–Clausen theorem. Hence, $\nu_3(S_a(b))=\nu_3(2a-1)$.

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ADDED. Explanation of the Faulhaber's formula implication:

Lemma. Let $n,m$ be positive integers, $m$ is even. Then $$1^{m} + 2^{m} + \dots + (n-1)^{m} \equiv B_{m} n \pmod{n}.$$

Proof. For $m=2$, the statement can be verified directly: $$1^2 + 2^2 + \dots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6} \equiv B_2 n\pmod{n}.$$ For the rest assume $m\geq 4$.

Using Faulhaber's formula and taking into account that $B_{m-1}=0$, we have \begin{split} & 1^{m} + 2^{m} + \dots + (n-1)^{m} \equiv 1^{m} + 2^{m} + \dots + n^{m} \\ &\equiv B_{m} n + \frac{1}{m+1} \sum_{k=3}^{m+1} \binom{m+1}{k} B_{m+1-k} n^k \pmod{n}. \end{split} It remains to show that $\frac{1}{m+1} \binom{m+1}{k} B_{m+1-k} n^k \equiv 0\pmod{n}$ for any integer $k\in [3,m+1]$.

Consider any prime $p\mid n$, and let $t:=\nu_p(n)\geq 1$. Our goal is to show that $\nu_p\big(\frac{1}{m+1} \binom{m+1}{k} B_{m+1-k}\big) + t(k-1) \geq 0$. It is enough to focus on the case $t=1$, from which the case $t>1$ follows instantly.

Since by von Staudt–Clausen theorem the denominators of Bernoulli numbers are square-free, we need to show that $$\nu_p\big(\frac{1}{m+1} \binom{m+1}{k}\big) + k - 2 \geq 0.$$

Noticing that $\nu_p\big(\frac{1}{m+1} \binom{m+1}{k}\big) = \nu_p\big(\frac1k\binom{m}{k-1}\big)\geq -\nu_p(k)$, our goal reduces to showing that $$k-2-\nu_p(k)\geq 0.$$ For $k=3$ and $k=4$, this inequality is trivial, while for $k\geq 5$ it follows from the bound $\nu_p(k)\leq \log_2(k)$. QED