Has the diophantine equation $ \sum_{n=1}^{N} \frac{1}{x_{n}} = \prod_{k=1}^{N} \left(1-\frac{1}{x_{k}} \right) $ been studied?

Question

Background

I wonder if there are any rational numbers such that their Egyptian fraction (sum) representations are equal to their Egyptian product analogue. In other words, I am curious about solutions to the diophantine equation

$$ \sum_{n=1}^{N} \frac{1}{x_{n}} = \prod_{k=1}^{N} \left(1-\frac{1}{x_{k}} \right) \tag{1}\label{1} $$

such that $x_{1} < x_{2} < \dots < x_{N-1} < x_{N} \in \mathbb{N}^{+}$ .

Solutions

• When $N=2$, the only solution appears to be $(x_{1},x_{2}) = (3,5) $, which yields the rational $\ \ $ number $\frac{8}{15}$.
• For $N=3$, we have the solution $(x_1, x_2, x_3) = (3,6,28)$, yielding the rational number $\frac{15}{28}$.
• For $N=4$, I've found eight solutions so far. Let $x_{a}^{4}$ be the $a$'th solution of the equation in four variables. We then have $x_{1}^{4} = (3, 7, 24, 52)$, $x_{2}^{4} = (3, 9, 15, 37) $, $x_{3}^{4} = (3, 9, 16, 32)$, $x_{4}^{4} = (3, 10, 15, 26)$, $x_{5}^{4} = (4, 5, 15, 36)$, $x_{6}^{4} = (4,5,17,28)$, $x_{7}^{4} = (4,7,8,35)$, and $x_{8}^{4} = (5,6,10,12)$. These solutions correspond to the rational numbers $\frac{391}{728}$, $\frac{899}{1665}$, $\frac{155}{288}$, $\frac{7}{13}$, $\frac{49}{90}$, $\frac{324}{595}$, $\frac{153}{280}$, and $\frac{11}{20} $, respectively.
• For $N \geq 5$, I haven't found any solutions yet. I suspect there are no solutions for the equation with this many variables -- though I'd gladly be proven wrong.

Related Equation

I know that in Znám's problem, solutions to the equation $$ \sum \frac{1}{x_{i}} + \prod \frac{1}{x_{i}} = y \tag{2}\label{2} $$ are studied, where $y$ and each $x_{i}$ must be integer. However, this equation is different from the one described above.

Question:

Has equation \eqref{1} been described and studied in the mathematical literature before?

Answer 1

I don't know if it has been studied before, but here is a partial result.

Claim : For all $N \geqslant 1$, there is a finite number of solutions.

Indeed, let $N \geqslant 1$, $(x_k)_{1 \leqslant k \leqslant N}$ a solution and $1 \leqslant n \leqslant N$ an integer. Let us give an upper bound on $x_n$ in function of $(x_k)_{1 \leqslant k \leqslant n - 1}$. We have, $$ \sum_{k = 1}^N \frac{1}{x_k} \leqslant \sum_{k = 1}^{n - 1} \frac{1}{x_k} + \frac{N - n + 1}{x_n}, $$ and, $$ \prod_{k = 1}^N \left(1 - \frac{1}{x_k}\right) \geqslant \prod_{k = 1}^{n - 1} \left(1 - \frac{1}{x_k}\right)\left(1 - \frac{1}{x_n}\right)^{N - n + 1}. $$ Therefore, if we set $S_n = \sum_{k = 1}^{n - 1} \frac{1}{x_k}$ ($= 0$ if $n = 1$) and $P_n = \prod_{k = 1}^{n - 1} \left(1 - \frac{1}{x_k}\right)$ ($= 1$ if $n = 0$), then, $$ P_n\left(1 - \frac{1}{x_n}\right)^{N - n + 1} \leqslant S_n + \frac{N - n + 1}{x_n} \qquad (1) $$ Notice also that, $$ \sum_{k = 1}^N \frac{1}{x_k} > S_n, \quad \prod_{k = 1}^N \left(1 - \frac{1}{x_k}\right) < P_n, \qquad (2) $$ hence $S_n < P_n$. For all positive real numbers $s,p,a$ such that $s < p$, let, $$ f_{s,p,a} : x \mapsto p\left(1 - \frac{1}{x}\right)^a - \frac{a}{x} - s. $$ We have $f_{s,p,a}(x) \rightarrow p - s > 0$ when $x \rightarrow +\infty$ so there exists some $x_{s,p,a} > 0$ large such that for all $x > x_{s,p,a}$, $f_{s,p,a}(x) > 0$. By point $(2)$, $S_n < P_n$ and by point $(1)$, $f_{S_n,P_n,N - n + 1}(x_n) \leqslant 0$ thus $x_n \leqslant x_{S_n,P_n,N - n + 1}$.

Let us show now that for all $1 \leqslant n \leqslant N$, there exists a universal bound $B_{n,N}$ that only depends on $n$ and $N$ such that $x_n \leqslant B_{n,N}$.

$n = 1 :$ $S_1 = 0$, $P_1 = 1$ and $N - n + 1 = N$ so we must have $x_1 \leqslant x_{0,1,N}$. $B_{1,N} = x_{0,1,N}$ fits.

$n \geqslant 2 :$ By induction, we have $x_k \leqslant B_{k,N}$ for all $k \leqslant n - 1$ and $S_n < P_n$. Let, $$ B_{n,N} = \max\{x_{S_n,P_n,N - n + 1}|\forall 1 \leqslant k \leqslant n - 1, x_k \leqslant B_{k,N} \textrm{ such that } S_n < P_n\}. $$ This only depends on $n$ and $N$ and this is well defined because the $x_k$ for $k \leqslant n - 1$ browse a finite set. And we necessarily have $x_n \leqslant B_{n,N}$ if $(x_k)_{1 \leqslant k \leqslant N}$ is a solution.

Everything in the proof is more or less explicit so if you can compute (or at least give a bound on) the $x_{s,p,a}$, then you can write a programm that computes all the solutions of function on $N$. However, it may take a lot of time when $N$ becomes too large.

Answer 2

Here's another partial result -- see my answer to OP's parallel question on Math Overflow for more details: for any N there is at least one solution, which we can construct as follows:

Write $p_1,q_1=2,1$. Then for $n\ge1$ let $p_{n+1},q_{n+1}=p_n^2+q_n,(p_n+1)q_n$. Then for any $n$ a solution to the equation here is $\{\,p_1+1,p_2+1,\ldots,p_n+1,p_{n+1}\,\}$. (And if we take all the $p+1$ then we get a solution to the "$N=\infty$" equation.)