Solving for the unknown in an exponential equation

Question

This is the equation: $(x^2 -x - 1)^{x^2}= x^2 -x - 1$.

I got $+1$ and $-1$ as the answers...but apparently even $2$ is a possible solution. I am not sure how to get it.

Answer 1

Consider the equation $a^b = a$. It's easy to verify that this has no solutions for $a,b \notin \{0,1\}$. Checking the remaining cases, we will have solutions for $a = 0$, $b\ne 0$; $a = 1$; and $b = 1$. So, letting $a = x^2 - x - 1$ and $b = x^2$, we solve these cases for $x$:

1. $a = 0$ gives $x^2 - x - 1 = 0$, which solves to $x = (1\pm \sqrt{5})/2$. Each of these gives a nonzero $b$.
2. $a = 1$ gives $x^2 - x - 1 = 1$, which solves to $x = -1, 2$
3. $b = 1$ gives $x^2 = 1$, which solves to $x = \pm1$

So the full solution set is $2, \pm 1, (1\pm\sqrt{5})/2$.