How many solutions are there for this equation: $(x^2-x-1)^{x^2}=(x^2-x-1)$ 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.
