Integer solution

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 $α = n + 2$.

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:

$$ 16 y^4 - 8(4n^2-4n+5)y^2 + 16n^4 - 32 n^3 - 8 n^2 + 24 n + 9 = 0 $$

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 $\alpha=1-n$.