<s>[INST] Presentations of rational numbers into product Prove or disprove:
For every rational number $q$ there exist rational numbers $q_1$, $q_2$, $q_3$ and $q_4$, such that:


*

*$q_1\times q_2\times q_3\times q_4 = q$

*$q_1+q_2+q_3+q_4 = 0$


My work so far:
I was able to solve it for $q_1$, $q_2$, $q_3$, $q_4$ and $q_5$:


*

*$-q\times\frac{q}{2}\times\frac{q}{2}\times\frac{2}{q}\times\left(-\frac{2}{q}\right)=q$

*$-q+\frac{q}{2}+\frac{q}{2}+\frac{2}{q}+\left(-\frac{2}{q}\right)=0$

 [/INST] It is a terrible solution (I obtained it by solving some diophantine equations):
$$q_1=\frac{(q-2^{10}\cdot3^{4})^2}{2^7\cdot3^4\cdot(q+2^9\cdot3^4)};$$
$$q_2=-4-\frac{q}{2^7\cdot3^4};$$
$$q_3 = \frac{2^3\cdot3\cdot(q+2^9\cdot3^4)}{q-2^{10}\cdot3^4};$$
$$q_4 = -\frac{2^{11}\cdot3^7\cdot q}{(q-2^{10}\cdot3^4)(q+2^{9}\cdot3^4)},$$
but it is a solution.
If $q = -2^{9}\cdot3^4$ or $q= 2^{10}\cdot3^4$ we shell obtain the decomposution for $\frac{q}{2^{12}\cdot 3^{12}}$ and after this multiplay $q_i$ by $2^{3}\cdot 3^{3}$.
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