Problem related to divisibility of even power sum The question was posted in MSE(12/19/20)link, but gets no answer. Hence I'm posting in MO
Define $S_m(n)=1^m+2^m+\cdots+n^m$

Can it be shown that
$S_{2m}(uv)\equiv0\pmod{uv}\iff S_{2m}(u)\equiv0\pmod{u}$ and $S_{2m}(v)\equiv0\pmod{v}$
Where $m,u,v$ are positive integer

Example:$1|S_2(1)=1,5|S_2(5)=55,7|S_2(7)=140,25|S_2(25)=5525,35|S_2(35)=14910,49|S_2(49)=40425$.
(1) if $p$ prime, $p\nmid S_{2m}(p)$ then $p-1\mid 2m$
Definition : For $m ≥ 0$, the $m$th power-sum denominator is the smallest positive integer $d_m$ such that $d_m · S_m(n)$ is a polynomial in $n$ with integer
coefficients.
The first few values of $d_m$ (see Sequence A064538) are
$d_m= 1, 2, 6, 4, 30, 12, 42, 24, 90, 20, 66, 24, 2730, 420, 90, 48, 510, . . . .$
(2) Clearly $\gcd(u,d_{2m})=1\implies u\mid S_{2m}(u)$.
(3) $p\mid d_{m}\implies p\le m+1$
I hope my following claim helps here:
• Consider $n\equiv 1\pmod2$ then $S_{2m}(n)\equiv0\pmod{n}\iff S_{2m}(\frac{n-1}2)\equiv0\pmod{n}$
Source code Pari/GP
for(m=1,30,for(a=2,100,if(sum(i=1,a,i^(2*m))%a==0,print([2*m,a,sum(i=1,a,i^(2*m))]))))

Proof for (1)
Proof for (3), see theorem 1 in this paper
 A: If $a,b$ are coprimes then
$\color{cadetblue}{a|S_{2m}(a)\ \text{and} \ b|S_{2m}(b)\ \text{iff} \ ab|S_{2m}(ab)}$.....$(1)$
$\mathcal Proof:$ Since, $S_{2m}(ab)=bS_{2m}(a) (\text{mod} \ a)$ and $S_{2m}(ab)=aS_{2m}(b) (\text{mod} \ b)$ and $(a,b)=1$, then surely $a|S_{2m}(a)$ and $b|S_{2m}(b)$ If $ab|S_{2m}(ab)$.
Conversely, if $a|S_{2m}(a)$ and $b|S_{2m}(b)$ and $S_{2m}(ab) \equiv r_{ab} (\text{mod} \ ab)$, then $a|r , b|r \Rightarrow lcm(a,b)=ab|r$ ( as $a,b$ are coprime) $\Rightarrow ab|S_{2m}(ab)$
Now, $$S_{2m}(p^{k+1})=\sum_{n=0}^{p-1}\sum_{t=1}^{p^k} (np^k+t)^{2m}$$
$\Rightarrow S_{2m}(p^{k+1}) \equiv pS_{2m}(p^k)+(\sum_{t=1}^{p^k} t^{2m-1})[(2m)\frac{p(p-1)}{2}p^k] \equiv pS_{2m}(p^k) (\text{mod}\ p^{k+1})$.
Hence, $\color{cadetblue}{p^{k+1}|S_{2m}(p^{k+1})\ \text{iff} \ p^{k}|S_{2m}(p^k)}$....$(2)$
Let,$uv=\prod {p_i}^{a_i}$ and $uv|S_{2m}(uv)$, then definitely , ${p_i}^{a_i}|S_{2m}({p_i}^{a_i})$ from $(1)$.Then $(2)$ implies $p_i|S_{2m}(p_i)$ and ${p_i}^{k}|S_{2m}({p_i}^{k})$ for any, $k$. As these all are different primes, we get $u|S_{2m}(u)$ and $v|S_{2m}(v) \cdots \text{qed}$.
