 Solution

For an integer $n\ge 0$, choose the series whose integer parity agrees with $n$ and set the other initial coefficient to zero. At $m=n$, the numerator in the recurrence vanishes, so $a_{n+2}=0$ and every later coefficient in that integer parity vanishes. With a nonzero coefficient of $x^n$, this gives a polynomial of degree exactly $n$. Normalize it by $P_n(1)=1$.

For $n=0$, the even branch is constant; for $n=1$, the odd branch is proportional to $x$. For $n=2$, the recurrence gives $a_2=-3a_0$, so normalization fixes $a_0=-1/2$. HenceThe opposite-parity branch generally does not terminate; polynomial termination does not mean that both independent solutions become polynomials.