Integer solution 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 $$(1-\mu^{2})\frac{d^{2}{\cal G}_{n}}{d\mu^{2}}+n(n-1){\cal G}_{n}=0$$
 A: From the observation by user35202, the equation actually can be factorized as$$
(α + n - 1)(α + n - 3)(α - n)(α - n - 2) = 0.
$$
So all the solutions are $α = -n + 1$, $α = -n + 3$, $α = n$, and $α = n + 2$.
A: 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:
$$
16 y^4 - 8(4n^2-4n+5)y^2 + 16n^4 - 32 n^3 - 8 n^2 + 24 n + 9 = 0
$$
A: Your recurrence relation seems to be wrong if you are using the differential equations you give. It should be instead $$\alpha(\alpha-1)(\alpha-2)(\alpha-3)-2n(n-1)(\alpha-1)(\alpha-2)+[n(n-1)-2]n(n-1)=0$$
One solution to this equation is $\alpha=n+2$. Another is $\alpha=3-n$. As pointed by Alex Francisco the other 2 solutions are $\alpha=n$ and $\alpha=1-n$.
A: Yes, there is a systematic way, but you might want to enlist computer help in that regard.
So we have the polynomial $$p(\alpha)=\alpha(\alpha-1)(\alpha-2)(\alpha-3)-4n(n-1)(\alpha-1)^2+n(n-1)[n(n-1)-2]$$ and we want integers $\alpha_i$ so that $p(\alpha_i)=0$ for integral $n$.
Note that in canonical form, the constant term of $p$ is given by $$n(n-1)[n(n-1)-2]-4n(n-1)=n(n-1)[n(n-1)-6]=k(n).$$ Now by the integral root corollary of the rational root theorem, all the integral roots $\alpha_i$ of $p$ must divide $k$. Thus we can find all the integral roots (when they exist) for arbitrary $n$; it suffices to check each of the factors of $k$ (for large $k$ this is obviously better by computer*) to see which is a root of $p$. Now this is always possible in a finite number of steps because every integer save $0$ has only a finite number of factors. As for $0$, we have $k=0$ only when $n=0$ or $n=1$ (since $k$ increases with $n$), and in these cases, the roots are clearly in $\{0,1,2,3\}$. This parity of roots, when they exist, is true in general for the integral pairs $r$ and $1-r$ when $r\ge2$. This clearly exhausts the integers $n$.
*Note that even by computer, some very large numbers cannot be readily factored, but this is only a technical problem; in theory, every integer can be expressed as a product of primes (disregarding the order of the factors).
