= Solution
Seek a <Frobenius solution> $y=z^r\sum_{n\geq0}a_nz^n$ with $a_0\ne0$. Since $(z+z^3)^2=z^2+2z^4+z^6$, its lowest coefficient gives the <indicial equation>
$$
[r(r-1)+3r+1]a_0=(r+1)^2a_0=0.
$$
Thus the repeated <indicial root> is $\boxed{r=-1}$. Equating the coefficient of $z^{n+r}$ before setting $r=-1$ gives
$$
(n+r+1)^2a_n+2(n+r-2)(n+r-3)a_{n-2}+(n+r-4)(n+r-5)a_{n-4}=0.
$$
For $n\geq1$ this becomes the <linear recurrence relation>
$$
\boxed{a_n=-\frac{2(n-3)(n-4)a_{n-2}+(n-5)(n-6)a_{n-4}}{n^2},\qquad a_j=0\text{ for }j<0.}
$$
Every coefficient is fixed by $a_0$, since the denominator never vanishes for $n\geq1$. In particular $a_1=0$ and then every odd coefficient is zero. Therefore there is $\boxed{\text{exactly one independent pure Frobenius series}}$, not two merely because the indicial root repeats. If "power series" means an ordinary <Taylor series> with nonnegative integer powers, the only such solution is zero: its coefficient equation starts with $a_0=0$ and inductively forces all coefficients to vanish.
A second independent local solution is a <logarithmic solution from a repeated Frobenius exponent>. Indeed, <Abel's identity> gives $W=Cz^{-3}[1+O(z^2)]$. For the normalized first solution $y_1=z^{-1}[1+O(z^2)]$, <reduction of order> gives
$$
y_2=y_1\int^z\frac{\exp[-\int^sP(v)dv]}{y_1(s)^2}\,ds.
$$
After choosing the harmless multiplicative constant in the numerator, its integrand is $1/s+O(s)$, so $y_2=y_1\log z+z^{-1}\times\text{an analytic series}$. The nonzero logarithmic coefficient proves that this second solution is not another pure <Frobenius solution>.
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