How many integers x satisfy the equation (x2 − x − 1)(x + 2) = 1?

A power a raised to b equals 1 in exactly three cases: (i) the base a = 1, for any exponent b; (ii) the base a = −1, when the exponent b is an even integer; or (iii) the exponent b = 0, provided the base a is not zero. For this question a = x2 − x − 1 and b = x + 2, and every case must be checked because x is required to be an integer.

1. Case: base = 1. Then x2 − x − 1 = 1, so x2 − x − 2 = 0, which factors as (x − 2)(x + 1) = 0, giving x = 2 or x = −1. The exponent is unrestricted here since 1 raised to any power is 1, so both values are valid.

2. Case: base = −1 with an even exponent. Then x2 − x − 1 = −1, so x2 − x = 0, giving x(x − 1) = 0, so x = 0 or x = 1. Check the exponent (x + 2) for each: at x = 0 the exponent is 2 (even), so (−1)2 = 1, valid. At x = 1 the exponent is 3 (odd), so (−1)3 = −1, not 1, so x = 1 is rejected.

3. Case: exponent = 0 with a non-zero base. Then x + 2 = 0, so x = −2. Check the base at x = −2: it equals 4 + 2 − 1 = 5, which is not zero, so the base is valid and 50 = 1 holds.

Collecting every case gives the distinct integer solutions. Verify each by direct substitution:

• At x = 2: base = 4 − 2 − 1 = 1, and 14 = 1, confirmed.

• At x = −1: base = 1 + 1 − 1 = 1, and 11 = 1, confirmed.

• At x = 0: base = −1, exponent = 2, and (−1)2 = 1, confirmed.

• At x = −2: base = 5, exponent = 0, and 50 = 1, confirmed.

No other integer satisfies any of the three cases, so the number of integer solutions is 4.