Let $n \in \mathbb{N}$ be a positive integer.

It is possible to prove that the only solution to the equation $F_1(n)=m$ with $m \in \mathbb{Z}$ and $$F_1(n)=(1+2+\dots+n)\cdot(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n})$$ is when $n=3$ and $m=11$. I've tried to generalize it to the case $$F_k(n)=(1^k+2^k+\dots+n^k)\cdot(\frac{1}{1^k}+\frac{1}{2^k}+\dots+\frac{1}{n^k})$$ for $k \geq 2$, i.e. asking for which $n$ the expression $F_k(n)$ is an integer, but I'm not succeeding. Some of you knows if this is possible to solve, for example where $k=2$?

I've tried with the basic elementary number theory techniques. Thanks in advance.

Let $n \in \mathbb{N}$ be a positive integer.

It is possible to prove that the equation $F_1(n)=m$ where $m \in \mathbb{Z}$ and $$ F_1(n)=(1+2+\ldots+n)\cdot\left(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}\right) $$ is solvable only for $n=3$ and $m=11$. I've tried to generalize the result for the function $$ F_k(n)=\big(1^k+2^k+\ldots+n^k\big)\cdot\left(\frac{1}{1^k}+\frac{1}{2^k}+\dots+\frac{1}{n^k}\right) $$ where $k \geq 2$, i.e. asking for which $n$ the expression $F_k(n)$ is an integer, but I'm not succeeding. Does some of you know if this equation is solvable in general or for some particular values of $k$, for example for $k=2$?

I've tried with the basic elementary number theory techniques. Thanks in advance.