The conversation between human and AI assistant.
How many values of [math]x[/math] ([math]x \in \mathbb{Z}[/math]) are there such that [math](x ^ 2 - x - 1) ^ {x + 2} = 1[/math]?
How did you arrive at that solution?
Well, first we notice that [math](x ^ 2 - x - 1) [/math] is always either [math]1[/math] or [math]-1[/math]. We can see this by noticing that [math](x ^ 2 - x - 1) = (x - \phi)(x - \psi)[/math], where [math]\phi[/math] and [math]\psi[/math] are the roots of the polynomial [math]x^2 - x - 1[/math]. Since [math]\phi[/math] and [math]\psi[/math] are the golden ratio and its inverse, we know that [math]\phi ^ 2 = \phi + 1[/math] and [math]\psi ^ 2 = \psi + 1[/math]. Using these equations, we can see that [math](\phi - 1)(\psi - 1) = -1[/math]. Therefore, [math](x ^ 2 - x - 1)[/math] is equal to [math]1[/math] when [math]x = -1[/math] and [math]x = 0[/math], and equal to [math]-1[/math] when [math]x = 1[/math]. Since [math]1 ^ {x+2}[/math] is always equal to 1, the equation is only satisfied for those three values of [math]x[/math].