<s>[INST] Divisibility of (finite) power sum of integers 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? $\nu_3(S_a(b))=\nu_3(2a-1)$.

 [/INST] 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, $\nu_3(S_a(b))=\nu_3(2a-1)$.

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
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