= Solution
If $N=|n|$ is an integer, the recurrence has zero numerator at $j=N$. Choosing the parity containing $a_N$ therefore makes all later coefficients vanish. Conversely, for a nonzero polynomial of degree $N$, the coefficient of $x^N$ in the differential equation is $(n^2-N^2)a_N$, so $n^2=N^2$ is necessary.
Normalizing the terminating series by $y(1)=1$ gives the first <Chebyshev polynomials>:
$$
\boxed{T_0(x)=1,\qquad T_1(x)=x,\qquad T_2(x)=2x^2-1,\qquad T_3(x)=4x^3-3x.}
$$
Substitution verifies each differential equation and endpoint value. More generally, $T_N(x)=\cos(N\arccos x)$; its recurrence $T_{N+1}=2xT_N-T_{N-1}$ also establishes polynomiality and the normalization.
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