Let $f(x)=c(x-a)(x-b)(x+a)(x+b)$ be a polynomial such that $a,b,c \in \mathbb{Q}$. Clearly the roots of this function are rational, and their sum is $0$. Their product is $a^2b^2$, which can be changed for any $x \in \mathbb{Q}$ to give any rational number. $\blacksquare$

S.Das93 wrote: Let $f(x)=c(x-a)(x-b)(x+a)(x+b)$ be a polynomial such that $a,b,c \in \mathbb{Q}$. Clearly the roots of this function are rational, and their sum is $0$. Their product is $a^2b^2$, which can be changed for any $x \in \mathbb{Q}$ to give any rational number. This doesn't make any sense. How do you even use $f(x)$ here? Not to mention that $a^2b^2$ only gives you the rational squares...

bump bump

My idea is consider a rational number $a$, we need to prove that there exist $x,y,z \in \mathbb{Q}$ such that \[xyz(x+y+z) = -a\]We can view this as a quadratic in $z$, so we need a rational discriminant, or \[(xy(x+y))^2 -4axy = t^2\] I don't know if this works, but if we take $xy = 4a$, we can write $t = 4au$ and simplify to \[(x+\frac{4a}{x})^2 - u^2 = 1\] now, $x^2-y^2= 1$ has rational point characterization $x = \frac{m^2+1}{m^2-1}$ and $y = \frac{2m}{m^2-1}$ for $m \in \mathbb{Q}$, so we want $x + \frac{4a}{x} = \frac{m^2+1}{m^2-1}$, or \[x - \frac{m^2+1}{m^2-1}x + 4a = 0\]to have a rational solution, i.e. \[\left(\frac{m^2+1}{m^2-1}\right)^2 - 16a = l^2\]for some rational $l$. idk if such $m$ and $l$ always exist, but it's just idea