How many solutions?

There are 3 ways $a^{b}$ can equal 1 when a and b are integers: I) a = 1; II) a = -1, b even; III) b = 0, a ≠ 0. In this problem $a = x^2-x -1, b = x +2$ . We solve each case in turn.

Case I:

$x^2 - x - 1 = 1, (x-2)(x+1) =0$ , $x=2$ or $-1$ .

Case II: $x^2-2- 1=-1$ and $x+2$ is even $x^2 - x = 0$ ,

$x=0$ or $1$ .

The choice $x = 0$ is a solution, since then $x +2$ is even. However, $x = 1$ is not a solution, since then $x + 2$ would be odd.

Case III:

$x +2 = 0$ and $x^2- x - 1 ≠ 0, x = -2$ . This is a solution, since $(-2)^2-(-2) - 1 = 5 ≠ 0$ . Thus there are 4 solutions in all.

$y=x+2$ gives me ${ \left( { y }^{ 2 }-5y+5 \right) }^{ y }$ . $y=0$ yields $x=\pm 2$ . Making the trial assumption that $y=1$ and solving ${ y }^{ 2 }-5y+4$ yields $y\epsilon \left\{ 4,5 \right\}$ and $x\epsilon \left\{ -1,0 \right\}$ .

I didn't think of the possibility that the exponent would be even.

• x + 2 = 0 --> x = -2

• x^2 - x - 1 = 1 --> (x - 2)(x + 1) = 0 --> x = -1 , 2

• x^2 - x - 1 = -1 --> x(x-1) = 0 --> x = 0 , 1 but x = 1 --> (-1)^3 = -1

so there are 4 intergers. {-2 , -1 , 0 , 2}