A generalisation of the equation $n = ab + ac + bc$

Question

In a result I am currently studying (completely unrelated to number theory), I had to examine the solvability of the equation $n = ab+ac+bc$ where $n,a,b,c$ are positive integers $0 < a < b < c.$

As it turned out the set of numbers not expressible in the above way is finite.

Generalizing the equation to four variables and checking the solutions of the equation $n = abc+abd+acd+bcd$ for $0 < a < b < c < d$ I've noticed that it looks like there exists a number $n_0$ such that for $n > n_0$ $n$ is expressible as $abc+abd+acd+bcd.$ The fact that a similar pattern occurs for five variables motivates me to ask the following question:

Question. Given a positive integer $m$ is there a number $n_0$ such that every $n > n_0$ is expressible as $$n = x_1\cdots x_m\left(\frac{1}{x_1} + \cdots + \frac{1}{x_m}\right)$$ where $0 < x_1 < x_2 <\ldots < x_m$.

The question is way too much for my (non-existent) knowledge of number theory. Perhaps there is a known result regarding such equations or, it can be somehow inductively derived from the case $m = 3.$ Any pointers in this direction are appreciated!

Answer 1

Assume the equation $$ n=x_1x_2\ldots x_N\left(\frac{1}{x_1}+\frac{1}{x_2}+\ldots+\frac{1}{x_N}\right),\tag 1 $$ where $x_i\in\textbf{N}$, $i=1,2,\ldots,N$

I will find the number of solutions of (1) and I will show that exist infinite sequence of natural numbers $n$, such that (1) have always solution.

Write $$ x_1 x_2\ldots x_N=t,\tag 2 $$ then $$ n=\frac{t}{x_1}+\frac{t}{x_2}+\ldots+\frac{t}{x_N}.\tag 3 $$ Given $n,t\in\textbf{N}$, the number of solutions of (3) under (2) is $$ r_0(n,t)=\sum_{\begin{array}{cc} d_1|t\textrm{, }d_2|t\textrm{, } \ldots\textrm{ , }d_N|t\\ \frac{1}{d_1}+\frac{1}{d_2}+\ldots+\frac{1}{d_N}=\frac{n}{t}\\ d_1d_2 \ldots d_N=t \end{array}}1. $$ Since the sum is a divisor sum we can rewrite it as $$ r_0(n,t)=\sum_{\begin{array}{cc} d_1|t\textrm{, }d_2|t\textrm{, } \ldots\textrm{ , }d_N|t\\ d_1+d_2+\ldots+d_N=n\\ d_1d_2 \ldots d_N=t^{N-1} \end{array}}1.\tag 4 $$ Now if we left $t$ varies in $\textbf{N}$, we get from Cauchy inequality $$ \frac{d_1+d_2+\ldots +d_N}{N}\geq\sqrt[N]{d_1d_2\ldots d_N}\Leftrightarrow \frac{n}{N}\geq\sqrt[N]{t^{N-1}}\Leftrightarrow t\leq\left(\frac{n}{N}\right)^{N/(N-1)}. $$ Hence the number of solutions of (1) is $$ r(n)=\sum_{t=1}^{\left[\left(\frac{n}{N}\right)^{N/(N-1)}\right]}\left(\sum_{\begin{array}{cc} d_1|t\textrm{, }d_2|t\textrm{, } \ldots\textrm{ , }d_N|t\\ d_1+d_2+\ldots+d_N=n\\ d_1d_2 \ldots d_N=t^{N-1} \end{array}}1\right).\tag 5 $$ Now it is clear from (5) that if happens $d_1=d_2=\ldots=d_{N-1}=t$, then $d_N=1$. But this happens if $n$ is of the form $$ n_k=(N-1)k+1\textrm{, }k\in\textbf{N}.\tag 6 $$ Hence when $n$ is of the form (6) we have always solution.