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

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!

• 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. – Will Jagy Jul 26 '10 at 18:37
• If we allow equality, the $n_0$ have been conjectured in oeis.org/classic/A027565, which appears to be growing exponentially. – tdnoe Jul 26 '10 at 22:32
• 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... – Harry Altman Jul 11 '11 at 12:51
• 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. – GH from MO Jul 11 '11 at 20:37
• 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 $x^2 + y^2 + z^9.$ – Will Jagy Jul 11 '11 at 22:40

## 1 Answer

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.

• Are you just showing that there are infinitely many integers $n$ that can be represented? That fact is obvious. – Thomas Browning Dec 15 '20 at 20:32
• @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. – Nikos Bagis Dec 15 '20 at 20:37