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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?

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  • $\begingroup$ If you found one solution with $n$ summands, just replace some positive $x_i$ with two numbers of size $2x_i$ to obtain a solution with $n+1$ summands. And one solution with $2$ summands is given by $x_1=-1$, $x_2=1$. $\endgroup$ – Hagen von Eitzen May 1 at 15:40
  • $\begingroup$ Just to be clear, are you fixing $k$ beforehand? So it would read like: "Fix $k$ in $\mathbb{N}$. Then there are infinitely many solutions for ... where $x_i \in \{-1,-2,...,-k\}$". Same for $n$. I think there are a few interpretations of the question, could oyu make it clearer? $\endgroup$ – NazimJ May 1 at 15:41
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    $\begingroup$ Not sure this is clear. We have $\frac 12+\frac 12=1$, and $\frac 13+\frac 13+\frac 13=1$ and so on...taking $x_1=-1$ we get an example of your sum for each $n≥3$, no? $\endgroup$ – lulu May 1 at 15:41
  • $\begingroup$ @lulu but I am starting with $n$ fixed. $\endgroup$ – user8795 May 1 at 15:43
  • $\begingroup$ Please edit your post to ask a clear question. As it stands, we're just guessing what is fixed and what we are meant to solve for. $\endgroup$ – lulu May 1 at 15:46
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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."

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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.

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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.

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