1985 AHSME Problems/Problem 21

Problem

How many integers $x$ satisfy the equation $\left(x^2-x-1\right)^{x+2} = 1?$

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of these}$

Solution

We recall that for real numbers $a$ and $b$ , there are exactly $3$ ways in which we can have $a^b = 1$ , namely $a = 1$ ; $b = 0$ and $a \neq 0$ ; or $a = -1$ and $b$ is an even integer.

The first case therefore gives $\begin{align*}x^2-x-1 = 1 &\iff x^2-x-2 = 0 \\&\iff (x-2)(x+1) = 0 \\&\iff x = 2 \text{ or } x = -1.\end{align*}$

Similarly, the second case gives $x+2 = 0$ , i.e. $x = -2$ , and this indeed gives $x^2-x-1 = 4+2-1 = 5 \neq 0$ , so $x = -2$ is a further valid solution.

Lastly, for the third case, we have $\begin{align*}x^2-x-1 = -1 &\iff x^2-x = 0 \\&\iff x(x-1) = 0 \\&\iff x = 0 \text{ or } x = 1,\end{align*}$ but $x = 1$ would give $x+2 = 3$ , which is odd, whereas $x = 0$ gives $x+2 = 2$ , which is even. Therefore, this case gives only one further solution, namely $x = 0$ .

Accordingly, the possible values of $x = -2$ , $-1$ , $0$ , or $2$ , yielding a total of $\boxed{\text{(C)} \ 4}$ solutions.

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1985 AHSME Problems/Problem 21

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