2017/027) Find all integers n such that $(n^2-n-1)^{n+2}= 1$

There are 3 cases when
$x^y= 0$ when y = 0 or x =1 or x = -1 and y even

hence  $(n^2-n-1)^{n+2}= 1$

when
1) n+2 = 0 or n = -2
or

2) $n^2 -n -1 = 1$
or $n^2 - n -2 = 0$ or $(n-2)(n+1) = 0$ giving n = 2 or -1

or
3) $n^2 -n -1 = -1 $   and n+2  is even
in this case
$x = -1 =>n^2 -n -1 = -1$ $n^2-n=0$ so n= 0 or n =1. but n+2 need to be even so n= 0

so solution set n= 0 or 2 or -2 or -1