From Apostol's Calculus, Vol. II, Section 6.21 #3:
The Legendre equation can be written in the form $\left[(x^2 -1)y'\right]'-\alpha(\alpha+1)y=0\,,$ where $\alpha\in\mathbb R$. If $a, b, c$ are constants with $a>b$ and $4c+1>0$, show that a differential equation of the type $\left[(x-a)(x-b)y'\right]'-cy=0$ can be transformed to a Legendre equation by a change of variable of the form $x=At+B$ where $A>0$. Determine $A$ and $B$ interms of $a$ and $b$.
It is easy to determine $A=\frac {a-b} 2$, and $B=\frac {a+b} 2$ (which agrees with the answer provided by the book), and we can find that this yields the equation $\left[(A^2t^2-A^2)y'\right]'-cy=0\implies\left[(t^2-1)y'\right]'-\frac c {A^2} y =0$ So to finish the proof that this can be considered as a Legendre equation, we must show that $\exists \alpha\in \mathbb R$ such that $\alpha(\alpha+1)=\frac c {A^2}$, however in solving this we find that $\alpha = \frac 1 2 \left(-1 \pm \sqrt{1+4\frac c {A^2}}\right)$. The condition $4c+1>0$ is not enough to guarantee that $\alpha\in\mathbb R$. For instance, we could take $a=\frac 1 4$, $b=0$, and $c=-\frac 1 8$.
Did I make a mistake, or should the condition instead be $16c + (a-b)^2>0$?