= Solution
{c}
Put $N=n(n+1)$. After division by $1-x^2$, the <Legendre differential equation> has coefficients
$$
p(x)=\frac{-2x}{1-x^2},\qquad q(x)=\frac{N}{1-x^2}.
$$
Both are <analytic functions> at zero, so $\boxed{x=0\text{ is an ordinary point}}$. At either $x_0=\pm1$, $p$ has a genuine <simple pole>; hence these are singular points even when $n=0$ makes $q$ vanish. However,
$$
\lim_{x\to x_0}(x-x_0)p(x)=1,\qquad
\lim_{x\to x_0}(x-x_0)^2q(x)=0,
$$
and both products extend analytically across $x_0$. By the <regular singular point criterion for a second-order equation>,
$$
\boxed{x=-1\text{ and }x=1\text{ are regular singular points}}.
$$
None of the three specified points is an <irregular singular point>.
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