My books says the possible solutions to $\hspace{0.2cm}$$(x^2-x-1)^{x^2}=(x^2-x-1)$ $\hspace{0.2cm}$in $\hspace{0.1cm}$$\mathbb{R}$ $\hspace{0.1cm}$ are $\hspace{0.1cm}$ $-1,1,2$
Is not $\hspace{0.2cm}$ $\frac{1+\sqrt5}{2}$ $\hspace{0.2cm}$ also a possible solution?
Plugging in the roots of $x^2-x-1$ yields $0^a=0$, which is true as long as $a\ne0$. Since neither of the roots of $x^2-x-1$ are also roots of $x^2$, they both satisfy the equation.
If $\displaystyle x^2-x-1=0, x=\cdots$
Else $\displaystyle(x^2-x-1)^{(x^2-1)}=1$
Now $\displaystyle a^m=1\implies$
either $m=0, a\ne0$
or $a=1$
or $a=-1,m$ even
