Is the following result true? Or Is there any known result about fractions like this?

Question

Is the following result true? Or Is there any known result of fractions like this?

Let $n$ be fixed.

There are infinitely many integer solutions for $$\sum_{i=1}^n \frac{1}{x_i} = 0,$$ where $x_i \in \Bbb{N}\ \cup \ \{-1,-2, \cdots, -k\}$, for some fixed $k$.

Here $\Bbb N$ is the set of natural numbers without $0$.

What about if all the $x_i's$ are distinct?

Also is there any pattern in the solutions?

Answer 1

If $k$ is fixed, and the $x_i$ distinct, then there are only finitely many possible values of the negative part, so if there are infinitely many solutions, there must be some positive number $x$ with infinitely many expressions as an Egyptian fraction.

If $n$ is fixed also, this is not possible, so the answer is "no."

Answer 2

To see that there are only finitely many solutions with $n,k$ fixed:

There are only finitely many multi-sets of size $≤n$ we can draw from $\{-1,\cdots, -k\}$. Let $s$ be one of these and let $S=\sum_{x_i\in s} \frac 1{x_i}$.

Now we want to argue that there are only finitely many multi-sets of size $≤n$ we can draw from $\mathbb N$ such that the sum of the reciprocals sums to a fixed value $N$ (in this case we want $N=-S$). If $n=1$ this is clear. We proceed by induction on $n$.

Take one such multi-set, call it $A$. Clearly we must have at least one $a\in A$ with $a≤\frac {|A|}N$ (else the sum of the reciprocals is too small). There are, of course, only finitely many such $a$. Remove this $a$ from $A$, and we now have a multiset of size $|A|-1≤n-1$ such that the sum of the reciprocals is $N-\frac 1a$. Inductively, we know there are only finitely many such so we are done.

Answer 3

No for fixed $n$ and $k$ there is only a finite number of solutions.

Indeed, let us try to build a solution. Let first pick $x_1,\ldots, x_{\ell}$; $x_{i}$ negative for $i \le \ell$. There are at most $k^n$ such choices.

Now let us assume WLOG that $x_{\ell+1},x_{\ell+2},\ldots, x_n$ are all positive and nondcreasing; $x_j \le x_{j+1}$ for each $j > \ell$. So we may assume that $\sum_{i=1}^{j} \frac{1}{x_j}$ is strictly negative for each $j$ satisfying $\ell \le j < n$, and therefore $|\sum_{i=1}^{j} \frac{1}{x_j}|$ $\geq$ $($lcm$(|x_{\ell}|,\ldots, x_j))^{-1}$ and so $x_{j+1} \le$ $n$lcm$(|x_{\ell}|,\ldots, x_j)$ which implies that $x_{j+1} \le n \prod_{i=\ell+1}^j x_i$ for each $j$ satisfying $\ell \le j < n$ [and $x_{\ell+1} \le n|x_{\ell}|$. This gives a finite number of solutions namely no more than $k^n(kn)^{n^2 \log n}$ possible solutions.