Let $x = a$ be an integral root of the cubic function above such that we can write the quadratic function in the parameter $m$: $m^2 + (a^2 - a)m + (1 - a^3) = 0$; or $m = \frac{(a - a^2) \pm \sqrt{(a - a^2)^2 - 4(1)(1 - a^3)}}{2}$; or $m = \frac{(a - a^2) \pm \sqrt{a^4 + 2a^3 + a^2 - 4}}{2}$; or $m = \frac{(a - a^2) \pm \sqrt{a^2(a + 1)^2 - 4}}{2}$ (i) In order for $m$ to be integral, the discriminant in (i) needs to be a perfect square, or: $a^2(a + 1)^2 - 4 = N^2$; or $a^2(a + 1)^2 - N^2 = 4$; or $[a(a+1) + N][a(a+1) - N] = 4$ (ii) Since 4 has the positive divisors 1, 2, and 4, these can be checked against the LHS factors in (ii): $a(a+1) + N = 4, a(a+1) - N = 1 \Rightarrow 2a(a+1) = 5$ (a contradiction & can be eliminated); $a(a+1) + N = 2, a(a+1) - N = 2 \Rightarrow 2a(a+1) = 4 \Rightarrow a(a+1) = 2 \Rightarrow a^2 + a - 2 = (a+2)(a-1) = 0 \Rightarrow a = -2, 1$ and $N = 0$. Finally, we compute the integral values of $m$ in (i) from our integral solutions for $a$ and $N$, which produces: $m = [(a - a^2) \pm N]/2$; $m = (1 - 1^2)/2 = 0$ and $(-2 - (-2)^2)/2 = -3$ or $\fbox{m = -3, 0}$.