Where does the restriction to integers comes from?
Taking $ (1 - x^2) u'' - 2 x u' + \nu u = 0 $ and $u = \sum_{k=0}^\infty a_k x^{k + s}$, the indicial equation is $s (s - 1) = 0$, which means only one solution can be generated by this ansat; $s = 0$ leads to the recurrence relation $ a_{k + 2} = \frac{k (k + 1) - \nu}{(k + 1) (k + 2)} a_k $ where $k \in \mathbb{Z}$, and here lies the (im)possibility of polynomial solutions.
The only way for $u$ to be a polynomial is that $ k (k + 1) - \nu = 0 $ has a solution in $\mathbb{Z}$.
The second solution is never a polynomial.
From MathWorld:
If $l$ is an integer, the function of the first kind reduces to a polynomial known as the Legendre polynomial.
If $l = \frac{\sqrt{13}-1}{2}$ the solution is not a polynomial.