We call an integer $k\geq 1$ good if for all $q\in\mathbb{Q}$ there are $a_1,\ldots, a_k\in \mathbb{Q}$ such that $$q = \prod_{i=1}^k a_i \cdot\big(\sum_{i=1}^k a_i\big).$$ Euler showed that $k=3$ is good.

Is the set of good positive integers infinite?


2 Answers 2


I suspect that $k = 4$ is good, but am not sure how to prove it. However, every positive integer $k \geq 5$ is good. This follows from the fact (see the proof of Theorem 1 from this preprint) that for any rational number $x$, there are rational numbers $a$, $b$, $c$, $d$ so that $a+b+c+d = 0$ and $abcd = x$. In particular, one can take $$ a(x) = \frac{2(1-4x)^{2}}{3(1+8x)}, b(x) = \frac{-(1+8x)}{6}, c(x) = \frac{-(1+8x)}{2(1-4x)}, d(x) = \frac{18x}{(1-4x)(1+8x)}, $$ as long as $x \not\in \{1/4, -1/8\}$. (For $x = 1/4$ one can take $(a,b,c,d) = (-1/2,1/2,-1,1)$ and for $x = -1/8$ one can take $(a,b,c,d) = (-2/3,25/12,-1/15,-27/20)$.)

Now, fix $k \geq 5$, let $q \in \mathbb{Q}$ and take $a_{1} = a(q/(k-4))$, $a_{2} = b(q/(k-4))$, $a_{3} = c(q/(k-4))$, $a_{4} = d(q/(k-4))$ and $a_{5} = a_{6} = \cdots = a_{k} = 1$. We have that $$ a_{1} + a_{2} + a_{3} + a_{4} + \cdots + a_{k} = 0 + a_{5} + \cdots + a_{k} = k-4 $$ and $a_{1} a_{2} a_{3} a_{4} \cdots = \frac{q}{k-4} \cdot 1 \cdot 1 \cdots 1 = \frac{q}{k-4}$. Thus $$ \left(\prod_{i=1}^{k} a_{i}\right) \left(\sum_{i=1}^{k} a_{i}\right) = \frac{q}{k-4} \cdot (k-4) = q. $$

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    $\begingroup$ For any rational number $x$, by Euler's result we can write $-x$ as $abc(a+b+c)$ with $a,b,c\in\mathbb Q$, thus $x=abcd$ with $d=-a-b-c$, and $a+b+c+d=0$. So Euler's result for $k=3$ essentially implies that $k\ge5$ is good. $\endgroup$ Jun 22, 2018 at 2:31
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    $\begingroup$ Indeed, we were unaware of Euler's result. $\endgroup$ Jul 20, 2018 at 0:06

Here I present the joint solution for $k=4$ by Guang-Liang Zhou and me. Actually, we not only show that $k=4$ is good, but also obtain a stroger result: Each rational number $q$ can be written as $abcd$ with $a,b,c,d\in\mathbb Q$ such that $a+b+c+d=1$. (We may also replace the number $1$ by any nonzero rational integer, but that's equivalent to the current version.)

Let $x=-81q/32$. If $q\not\in\{4/81,-8/81\}$, then $x\not\in\{1/4,-1/8\}$, hence $$a(x)+b(x)+c(x)+d(x)=1$$ and $$a(x)b(x)c(x)d(x)=-\frac{32}{81}x=q,$$ where \begin{gather*}a(x)=-\frac{(1+8x)^2}{9(1-4x)},\ \ b(x)=\frac49(1-4x), \\ c(x)=\frac{2(1-4x)}{3(1+8x)},\ \ d(x)=\frac{12x}{(1-4x)(1+8x)}.\end{gather*} For $q=4/81$, clearly $$\frac4{81}=\frac{4}3\times\frac13\times\left(-\frac13\right)\times\left(-\frac13\right)\ \ \text{with}\ \frac 43+\frac13+\left(-\frac13\right)+\left(-\frac13\right)=1.$$ Similarly, for $q=-8/81$ we have $$-\frac{8}{81}=\frac{1}3\times\frac23\times\frac23\times\left(-\frac23\right)\ \ \text{with}\ \frac 13+\frac23+\frac23+\left(-\frac23\right)=1.$$

  • $\begingroup$ Amazing, thank you! This answer would also deserve acceptance like the one above $\endgroup$ Jun 23, 2018 at 15:45

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