Solve $(a + \sqrt {a^2 - 1})^{x^2 - 2x} + (a - \sqrt {a^2 - 1})^{x^2 - 2x} - a = 0$ for $x$ , where $a>1$ .
My approach is as follows :
$(a + \sqrt {a^2 - 1}) (a - \sqrt {a^2 - 1})=1 $
Let $(a + \sqrt {a^2 - 1})^{x^2 - 2x}=y$. From here, I got stuck.
The answer (as given at the back of the book) is $1$, $1 + \sqrt{2}$, $1 - \sqrt{2}$.
I think it should be $(a + \sqrt {a^2 - 1})^{x^2 - 2x} + (a - \sqrt {a^2 - 1})^{x^2 - 2x} - 2a = 0$
So that we have
$$y+\dfrac1y=2a\iff y^2-2ay+1=0$$
$\implies y=\dfrac{2a\pm\sqrt{(2a)^2-4}}2=a\pm\sqrt{a^2-1}$
Case$\#1:$ if $y=a+\sqrt{a^2-1},$
$(a+\sqrt{a^2-1})^{x^2-2x}=a+\sqrt{a^2-1}$
Now if $p^x=p^y$
either $x=y$
or $p=1$
or $p=0, x,y>0$
or $p=-1, x-y=$even integer
Case$\#2:$ if $y=a-\sqrt{a^2-1},$
$(a+\sqrt{a^2-1})^{x^2-2x}=a-\sqrt{a^2-1}(a+\sqrt{a^2-1})^{-1}$
