TITLE: Question on a generalisation of a theorem by Euler
QUESTION [22 upvotes]: 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?

REPLY [30 votes]: 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
$$-\frac{8}{81}=\frac{1}3\times\frac23\times\frac23\times\left(-\frac23\right)\ \ \text{with}\ \frac 13+\frac23+\frac23+\left(-\frac23\right)=1.$$