Finding rational solutions to $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but $abc\notin\mathbb{Z}$ I'm trying to find $a,b,c\in\mathbb{Q}$ such that $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but that $abc\notin\mathbb{Z}$.
I tried to pick specific values for $a, b$
$$
b^3x^2 + a^3x - n= 0
$$
where I tried different integers $n\neq 0$ and then evaluate the roots of this polynomial. However, this proved unsuccessful and now I'm stuck.
 A: I don't know offhand of any way to use your approach with $b^3x^2 + a^3x - n= 0$. Instead, since you're looking for just one set of values, consider that $a = \frac{3}{2}$, $b = \frac{5}{2}$ and $c = 1$ gives $a^3c + b^3c^2 = \frac{3^3}{2^3} + \frac{5^3}{2^3} = \frac{27 + 125}{8} = 19 \in \mathbb{Z}$, but $abc = \frac{15}{4} \not\in \mathbb{Z}$.
In general, with $c = \pm 1$, then for any integers $d$, $e$ and $f$, where $\lvert f \rvert \gt 1$ and $\gcd(d,f) = \gcd(e,f) = 1$, plus there's a non-zero integer $n$ such that
$$cd^3 + e^3 = nf^3 \tag{1}\label{eq1A}$$
then $a = \frac{d}{f}$ and $b = \frac{e}{f}$ works since $abc = \frac{dec}{f^2} \not\in \mathbb{Z}$. For example, with $c = 1$, then $e = f^3 - d$ (or, more generally, $e = kf^3 - d$ for any non-zero integer $k$) works in \eqref{eq1A} since $d^3 + (f^3 - d)^3 = d^3 + f^9 - 3f^6d + 3f^3d^2 - d^3 = (f^6 - 3f^3d + 3d^2)f^3$, giving $n = f^6 - 3f^3d + 3d^2$. With $c = -1$, then $e = kf^3 + d$ (with $k \neq 0$) works instead.
There are undoubtedly other techniques to determine different groups of $(a,b,c)$ which also work, with mine likely among the simplest of these techniques.
A: COMMENT.-Certainly you do want to have $a,b,c\in\mathbb Q\setminus\mathbb Z$ so I am afraid you have affaire to very particular elliptic curves of the form $Ax^3+BY^3=CZ^3$ which have been seriously studied first by E. S. Selmer (The diophantine equation $ax^3+by^3+cz^3=0$ Act. Math. (Stockh.) 85 (1951) p.203-362). It is a topic in which the difficulty is very hard mainly if you want to impose restrictive conditions like yours.
In fact, put $\dfrac{a}{\alpha}, \dfrac{b}{\beta},\dfrac{c}{\gamma}$ your rational with $\dfrac{abc}{\alpha\beta\gamma}\notin\mathbb Z$ then your condition becomes
$$\gamma c(\beta a)^3+c^2(\alpha b)^3=n(\alpha\gamma)^3$$
However you have at hand some divisibility conditions you can maybe used to get divers class of solutions. In any case you do have to try with a Selmer curve.
