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? $\nu\_3(S\_a(b))=\nu\_3(2a-1)$.
> 
> 
> 



Divisibility of (finite) power sum of integers
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](https://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem).
Hence, $\nu\_3(S\_a(b))=\nu\_3(2a-1)$.




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**ADDED.** Explanation of the [Faulhaber's formula](https://en.wikipedia.org/wiki/Faulhaber%27s_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


