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](https://mathoverflow.net/a/268336/8628) that $k=3$ is good.


Is the set of good positive integers infinite?



Question on a generalisation of a theorem by Euler
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](https://arxiv.org/pdf/1607.01957.pdf)) 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
$$
 \left(\prod\_{i=1}^{k} a\_{i}\right) \left(\sum\_{i=1}^{k} a\_{i}\right) = \frac{q}{k-4} \cdot (k-4) = q.
$$


