Can $y^2-4$ be a divisor of $x^3-x^2-2 x+1$? Do there exist integers $x$ and $y$ such that $\frac{x^3-x^2-2 x+1}{y^2-4}$ is an integer?
In other words, can any integer representable as $x^3-x^2-2 x+1$ have any divisor representable as $y^2-4$?
This is the simplest non-trivial example of my earlier question Integer points of rational function in 2 variables .
 A: No.  The roots of $x^3 - x^2 - 2x + 1$ are $-(\zeta + \zeta^{-1})$ where $\zeta$ is a 7th root of unity; this soon implies [see below] that any prime factor is either $7$ or $\pm 1 \bmod 7$, and thus that all factors of $x^3 - x^2 - 2x + 1$ are congruent to $0$ or $\pm 1 \bmod 7$.  In particular it is not possible for two factors to differ by $4$, so no number of the form $y^2 - 4 = (y-2) (y+2)$ can divide $x^3 - x^2 - 2x + 1$.
added later: To show that any prime factor $p$ of $x^3 - x^2 - 2x + 1$ is either $7$ or $\pm 1 \bmod 7$, let $k$ be the finite field of order $p^2$, and $\zeta \in k$ a root of the quadratic equation $\zeta^2 + x\zeta + 1 = 0$ (any quadratic equation with coefficients in the $p$-element field has a root in $k$).  Then $\zeta^7 = 1$, so either $\zeta = 1$ or the multiplicative group $k^\times$ of $k$ has a subgroup of size $7$.  In the former case, $x = -2$, and then $x^3 - x^2 - 2x + 1 = -7$ so $p=7$.  In the latter case, Lagrange's theorem gives $7 \mid \#k^\times = p^2-1$, so $p \equiv \pm 1 \bmod 7$.  QED
