= Solution
Divide the <Chebyshev differential equation> by $1-x^2$. Its coefficients $-x/(1-x^2)$ and $m^2/(1-x^2)$ are analytic at zero, so $\boxed{x=0\text{ is an ordinary point}.}$ The finite <singular points of a differential equation> are $x=\pm1$.
Insert $y=\sum_{n\geq0}a_nx^{n+\sigma}$, $a_0\ne0$. The coefficient of the lowest power $x^{\sigma-2}$ is $\sigma(\sigma-1)a_0$, hence $\sigma=0$ or $1$. The next power gives $\sigma(\sigma+1)a_1=0$, and the remaining powers yield
$$
\boxed{a_{n+2}=\frac{(n+\sigma)^2-m^2}{(n+\sigma+2)(n+\sigma+1)}a_n,\qquad n\geq0.}
$$
For $\sigma=0$, $a_1$ is free and
$$
a_{n+2}=\frac{n^2-m^2}{(n+2)(n+1)}a_n.
$$
Choosing $a_1=0$ selects the even <power series>, with $y(0)=a_0,y'(0)=0$. For $\sigma=1$, the next-power equation forces $a_1=0$ and
$$
a_{n+2}=\frac{(n+1)^2-m^2}{(n+3)(n+2)}a_n.
$$
This selects the odd <power series>, with $y(0)=0,y'(0)=a_0$. Their different initial values prove <linear independence>. Although the <Frobenius method> produces two possible leading exponents, zero is an <ordinary point>, and a general $\sigma=0$ series already includes both initial constants.
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