If x is an integer, then which of the following statements about $x^2 - x - 1$ is true?

MCQ: If x is an integer, then which of the following statements about $x^2 - x - 1$ is true?

1. It is always odd.

2. It is even when x is even and odd when x is odd.

3. It is always even.

4. It is always positive.

The correct answer is: "It is always odd."

Reasoning

To determine why $x^2 - x - 1$ is always odd, let's examine the expression more closely by considering the parity (evenness or oddness) of x.

Parity of x

• If x is even: Let x = 2k, where k is an integer.

  • $x^2 = (2k)^2 = 4k^2$ (which is even)

  • $-x = -(2k) = -2k$ (which is even)

  • $-1$ (which is odd)

• When you sum these terms for an even x: <br>$x^2 - x - 1 = 4k^2 - 2k - 1$\br> Since $4k^2$ and $-2k$ are both even, the sum of two even numbers is even. Upon subtracting an odd number (-1), the final result is odd. Therefore, when x is even, $x^2 - x - 1$ is odd.

• If x is odd: Let x = 2k + 1, where k is an integer.

  • $x^2 = (2k + 1)^2 = 4k^2 + 4k + 1$ (which is odd)

  • $-x = -(2k + 1) = -2k - 1$ (which is odd)

  • $-1$ (which is odd)

• When you sum these terms for an odd x: <br>$x^2 - x - 1 = (4k^2 + 4k + 1) - (2k + 1) - 1$\br> Simplifying, we get: $4k^2 + 4k + 1 - 2k - 1 - 1 = 4k^2 + 2k - 1$ \br> The sum of odd terms $(1-2k)$ with any multiple of two is odd. Therefore, when x is odd, $x^2 - x - 1$ remains odd.

By analyzing both cases, we can conclude that the expression $x^2 - x - 1$ results in an odd number whether x is even or odd.

Conclusion:

Thus, the statement "It is always odd." is indeed the correct one among the given options. This conclusion is based on the proof of parity analysis which shows that the expression $x^2 - x - 1$ consistently results in an odd number for any integer value of x.