Prove that there are $a_{1},a_{2},...,a_{n}$ for which $a_{1}a_{2}...a_{n}(\frac{1}{a_{1}} +\frac{1}{a_{2}}+...+\frac{1}{a_{n}})$ is a perfect square. Prove that for any $n\geq3$ there are distinct nonzero natural numbers $a_{1},a_{2},...,a_{n}$ for which $a_{1}a_{2}...a_{n}(\frac{1}{a_{1}} +\frac{1}{a_{2}}+...+\frac{1}{a_{n}})$ is a perfect square.
My work so far: I've thought about the case of equality for Cauchy-Schwartz inequality ($a_{1}a_{2}...a_{n}(\frac{1}{a_{1}} +\frac{1}{a_{2}}+...+\frac{1}{a_{n}})\geq{n}^2$): $a_{1}^2=a_{2}^2=...=a_{n}^2$, but they are distinct numbers.
Any help is appreciated!
 A: Hint
An important feature of such problems is scalability. In this problem, assume you are given a set of distinct numbers $b_1,\cdots,b_n$. Now, define $a_i=kb_i$ for $i=1,2,\cdots,n$ and try to adjust $k$ such that $a_{1}a_{2}...a_{n}(\frac{1}{a_{1}} +\frac{1}{a_{2}}+...+\frac{1}{a_{n}})=k^{n-1}b_{1}b_{2}...b_{n}(\frac{1}{b_{1}} +\frac{1}{b_{2}}+...+\frac{1}{b_{n}})$ is a perfect square.
Update
Some folks suggested that this approach will not work for odd values of $n$. I hope the even $n$ case is clear enough and everyone can go through it. The solution for odd $n$ is as follows.
Fix an integer $r\ge 2$ and let $a_1=r(r+1)$ and $a_2=r+1$. Then
$$
a_1\cdots a_n\left(\frac{1}{a_1}+\cdots+\frac{1}{a_n}\right)=
r(r+1)^2a_3\cdots a_n\left(\frac{1}{r}+\frac{1}{a_3}+\cdots+\frac{1}{a_n}\right),
$$for which, a sufficient condition for fulfilling the OP's request is that
$$
a_1\cdots a_n\left(\frac{1}{a_1}+\cdots+\frac{1}{a_n}\right)=
ra_3\cdots a_n\left(\frac{1}{r}+\frac{1}{a_3}+\cdots+\frac{1}{a_n}\right)
$$ be a perfect square. Since we are dealing with an even number of components, the case of odd $n$ reduces to the case of even $n$ $\blacksquare$
A: Here is a strategy to find such numbers.
Let's assume $a_1=1, a_2=2,a_3=3,  ..., a_n=n, a_{n+1}=x, a_{n+2}=kx$. We are going to find a pair $(x,k)$ such that the sequence $a_1, ..., a_{n+2}$ satisfies the condition of the problem.
It is very easy to see that our goal, now,  is to make
$$x(n!+kx(\frac{n!}{1}+\frac{n!}{2}+ ... +\frac{n!}{n})+k(n!))$$
a perfect square. To do so, we may suppose:
$$n!+kx(\frac{n!}{1}+\frac{n!}{2}+ ... +\frac{n!}{n})+k(n!)=m^2x,$$
where $m$ is a positive integer. Therefore our goal, now, is to find a triple $(x,k,m)$ such that the equality above holds.
We have:
$$n!+kx(\frac{n!}{1}+\frac{n!}{2}+ ... +\frac{n!}{n})+k(n!)=m^2x \\ \iff x(m^2-k(\frac{n!}{1}+\frac{n!}{2}+ ... +\frac{n!}{n}))=n!+k(n!). $$
An example of such $(x,k,m)$, is:
$$m+1=\frac{n!}{1}+\frac{n!}{2}+ ... +\frac{n!}{n}\\
k=m-1 \\ x=n!+k(n!).$$
We are done.

Note: There was no obligation to assume $a_1=1, a_2=2,a_3=3,  ..., a_n=n$ as initial values. We just chose these simple values to have an explicit example.
For example, if $n+2=4$, put $a_1=2, a_2=3$, then by following the same argument as above, we have: $k=3$ and $x=24$. So $(a_1,a_2,a_3,a_4)=(2,3,24,72);$
and, if $n+2=5$, we can easily obtain $(a_1,a_2,a_3,a_4,a_5)=(1,2,3,60,540).$
