Power Series (Cambridge Tripos 1900) If 
$$a/(a+bz+cz^2)=1+p_1z+p_2z^2
+\dots$$
then
$$1+p_1^2z+p_2^2z^2+\dots=\frac{a+cz}{a-cz}\frac{a^2}{
a^2-(b^2-2ac)z+c^2z^2}$$
A tricky problem from G.H.Hardy's "A Course in Pure Mathematics", or maybe I'm missing the obvious.Any help will be greatly appreciated.

If we start with the original function, we can reformulate is conveniently as
$$
\frac{a}{a+bz+cz^2} = \frac{1}{1+(\frac{b}{a})z+(\frac{c}{a})z^2} = \frac{1}{(1-\frac{z}{z_+})(1-\frac{z}{z_-})}  = \frac{z_+}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_-})} - \frac{z_-}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_+})}
$$
with $z_\pm$ the roots of the denominator, i.e.,
$$
z_\pm = \frac{- b \pm \sqrt{b^2 - 4 a c}}{2 c}
$$
If we expand the two expressions on the left as a series in $z$, we get
$$
\frac{z_+}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_-})} = \sum_{n\geq0} l_n z^n \quad \text{ with } \quad l_n = \frac{z_+}{(z_+-z_-)} \frac{1}{z_-^n}
$$
$$
\frac{z_-}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_+})} = \sum_{n\geq0} r_n z^n \quad \text{ with } \quad r_n = \frac{z_-}{(z_+-z_-)} \frac{1}{z_+^n}
$$
and hence we identify $p_n=l_n-r_n$ for $n\geq0$. Now let us construct a series with coefficients $p_n^2 = (l_n-r_n)^2$. It is relatively easy to recognise that we can make the following identifications
$$
\sum_{n\geq0} l_n^2 z^n = \frac{z_+^2}{(z_+-z_-)^2} \frac{1}{(1-\frac{z}{z_-^2})}
$$
$$
\sum_{n\geq0} r_n^2 z^n = \frac{z_-^2}{(z_+-z_-)^2} \frac{1}{(1-\frac{z}{z_+^2})}
$$
$$
\sum_{n\geq0} l_n r_n z^n = \frac{z_+ z_-}{(z_+-z_-)^2} \frac{1}{(1-\frac{z}{z_+ z_-})}
$$
and hence that the required series
$$
\sum_{n\geq0} p_n^2 z^n = \frac{1}{(z_+-z_-)^2} \left(\frac{z_+^2}{(1-\frac{z}{z_-^2})} +  \frac{z_-^2}{(1-\frac{z}{z_+^2})} - \frac{2 z_+ z_-}{(1-\frac{z}{z_+ z_-})}\right) = 
\frac{z_+^2 z_-^2 (z_+ z_- + z)}{(z_+ z_- -z)(z_-^2 - z)(z_+^2 - z)}
$$
Using that $z_+ z_- = a/c$ and $z_+^2+z_-^2=(b^2-2 ac)/c^2$ it then follows that 
$$
\sum_{n\geq0} p_n^2 z^n = \frac{a+c z}{a - c z} \frac{a^2}{a^2 - (b^2-2ac)z + c^2 z^2}
$$
