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

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?

• 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 $x_1=-1$, $x_2=1$. – Hagen von Eitzen May 1 at 15:40
• 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? – NazimJ May 1 at 15:41
• 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? – lulu May 1 at 15:41
• @lulu but I am starting with $n$ fixed. – user8795 May 1 at 15:43
• 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. – lulu May 1 at 15:46

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."

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.

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.