likely open number theory problem: finite sum of $\zeta(2)$ equal to a square of rationals Which $n$ can let $S=1+\frac14+\frac19+\cdots+\frac1{n^2}$ be a square of a rational number? Obviously, $1$ and $3$ work, but how to prove they are the only ones?
I think this problem is really hard. I have asked many professional number theorists in a top-5 math department in US, and none of them can give any clue. I am thinking this could be an open question.
If you cannot solve this, try this: let $S=1+\frac12+\frac13+\cdots+\frac1{n}$. Which n can make this an integer. Prove your result. This is proposed by a famous mathematician Shing Tung Yau for a college math competition. 
Hopefully someone can give a clue to No.1. I suggested if you cannot solve the first one but would like to see an answer, rate this problem up so that it may have a better chance to bring in more experts for a solution. Thanks!
It is also appreciated if you would like to post your thoughts on Yau's problem!

The first problem, which essentially asks whether $\sqrt{H_n^{(2)}}$, where $H_n^{(2)}$ is a Generalized Harmonic Number, can ever be rational for $n\ge4$, is very hard. I don't see a way to approach it.

The Second Problem
Let $k_n\in\mathbb{Z}$ be so that $2^{k_n}\le n\lt2^{k_n+1}$. The only integer not exceeding $n$ divisible by $2^{k_n}$ is $2^{k_n}$.
Let $d_n=b_n2^{k_n}$ be the LCM of $\{k:1\le k\le n\}$, where $b_n$ is odd. $\dfrac{d_n}{2^{k_n}}$ is the only odd number in
$$
\left\{\dfrac{d_n}{k}:1\le k\le n\right\}
$$
Therefore,
$$
d_n\sum_{k=1}^n\frac1k
$$
is an odd number. Since $d_n$ is an even number for $n\ge2$, we get that
$$
\sum_{k=1}^n\frac1k
$$
cannot be an integer for $n\ge2$.
