I'm wondering whether the Legendre polynomials $P_m(x)$ and $P_{m+2k}(x)$, with $m$ even and $k \in \mathbb{N}^+$, can have common roots.

For $k=1$ it is straightforward to show that there are no common roots (see proof below) but I'm not sure how to prove (or disprove) it for $k > 1$. Perhaps by induction?

Given Bonnet’s recursion formula

$$(n+1) P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x),$$

it follows that consecutive Legendre polynomials can not have common roots. Indeed, if $P_{n+1}(x)=0$ and $P_n(x)=0$ it follows that $P_{n-1}(x)=0$. But then $P_{n-2}(x), \ldots, P_1(x)=0$. As $P_1(x)$ only has a root at $x=0$ and $P_2(0) \neq 0$ we have a contradiction.

Rewriting the relation as

$$(2n+1)xP_n(x) = (n+1)P_{n+1}(x) + nP_{n-1}(x),$$

we see that if $P_{n+1}(x)=0$ and $P_{n-1}(x)=0$ then $x=0$, because $P_n(x) \neq 0$ as proven above. As $P_n(0)=0$ for odd $n$ it follows that $P_{n+1}(0)\neq 0$ and $P_{n-1}(0)\neq 0$, so the Legendre polynomials $P_{n+1}(x)$ and $P_{n-1}(x)$ do not have common roots.



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