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!

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    $\begingroup$ I would write this as asking for a representation with all positive $x_j$ as $$ n = x_1 x_2 \ldots x_m \left( \frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_m} \right) $$ for any $ n > n_0.$ This agrees with your examples for $m=3$ and $m=4.$ Is this what you want? I find your way of writing this and speaking of a set $X$ as clouding the issue. $\endgroup$
    – Will Jagy
    Jul 26, 2010 at 18:37
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    $\begingroup$ If we allow equality, the $n_0$ have been conjectured in oeis.org/classic/A027565, which appears to be growing exponentially. $\endgroup$
    – tdnoe
    Jul 26, 2010 at 22:32
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    $\begingroup$ By "taken from a set", what you mean is that they're distinct, right? "Taken from a set" doesn't really mean anything. But it probably is not worth worrying about distinctness at first... $\endgroup$ Jul 11, 2011 at 12:51
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    $\begingroup$ This looks like a very difficult problem as the number of variables $m$ is almost the same as the degree $m-1$. It is of similar flavor to the question "Can we write every sufficiently large number as a sum of four cubes?" or "Is $G(k)<100k$ in the Waring problem?" I would be surprised if this problem were resolved in the next 20 years. $\endgroup$
    – GH from MO
    Jul 11, 2011 at 20:37
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    $\begingroup$ GH, I must agree, I was just illustrating that the rough density argument may not be enough. I remember, though, R.C. Vaughan telling Kaplansky that the obstruction here could not be detected $p$-adically, and as such this defeated a conjecture in the first edition of his book on the Hardy-Littlewood method. So, in the second edition, on page 127 it says "There are some exceptions to this, see Exercise 5," then Exercise 5 on page 146 is about $x^2 + y^2 + z^9.$ $\endgroup$
    – Will Jagy
    Jul 11, 2011 at 22:40

1 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.

  • $\begingroup$ Are you just showing that there are infinitely many integers $n$ that can be represented? That fact is obvious. $\endgroup$ Dec 15, 2020 at 20:32
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    $\begingroup$ @ThomasBrowning no mostly the representation part formula (5). I know that all integers of the form $n=t^{N-1}+(N-1)t^{N-2}$ are solutions. $\endgroup$ Dec 15, 2020 at 20:37

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