TITLE: Integer solution
QUESTION [8 upvotes]: I would like to solve the following equation for $\alpha$ under the condition that $\alpha$ is an integer (positive or negative)
$$\alpha(\alpha-1)(\alpha-2)(\alpha-3)-2n(n-1)(\alpha-1)(\alpha-2)+[n(n-1)-2]n(n-1)=0$$
where $n$ is an integer. Is there a systematic way/method to solve it or I just have to use trial and error?
For some background, this is a relation that follows from the differential equation
$$\alpha(\alpha-1)(\alpha-2)(\alpha-3){\cal G}_{n}+4(1-\mu^{2})\frac{d^{2}{\cal G}_{n}}{d\mu^{2}}-4(1-\mu^{2})\alpha \frac{d^{2}{\cal G}_{n}}{d\mu^{2}}+2(1-\mu^{2})\alpha(\alpha-1)\frac{d^{2}{\cal G}_{n}}{d\mu^{2}}+(1-\mu^{2})^{2}\frac{d^{4}{\cal G}_{n}}{d\mu^{4}}-4\mu(1-\mu^{2})\frac{d^{3}{\cal G}_{n}}{d \mu^{3}}=0
$$
where ${\cal G}_{n}$ obeys the equation $$(1-\mu^{2})\frac{d^{2}{\cal G}_{n}}{d\mu^{2}}+n(n-1){\cal G}_{n}=0$$

REPLY [5 votes]: For a more "systematic way" which doesn't rely on "seeing" the right factorization, consider exploiting the symmetry around $\,\alpha=\frac{3}{2}=\frac{0+3}{2}=\frac{1+2}{2}\,$. Let $x-3/2 = y$, then substituting $x = y + 3/2$ in the original equation gives a biquadratic in $y$ i.e. a quadratic in $y^2$, which can then be routinely solved in the general case, without any assumptions about integer values:
$$
16 y^4 - 8(4n^2-4n+5)y^2 + 16n^4 - 32 n^3 - 8 n^2 + 24 n + 9 = 0
$$