Question

Which of the following properly uses a polynomial identity to detect if 22k−1 is prime?(1 point)
Responses

22k−1=(22+1)(2k−1), so 22k−1 is not prime.
2 superscript 2 k baseline minus 1 equals left parenthesis 2 superscript 2 baseline plus 1 right parenthesis left parenthesis 2 superscript k baseline minus 1 right parenthesis , so 2 superscript 2 k baseline minus 1 is not prime.

22k−1=(2k+1)(2k−1), so 22k−1 is not prime.
2 superscript 2 k baseline minus 1 equals left parenthesis 2 superscript k baseline plus 1 right parenthesis left parenthesis 2 superscript k baseline minus 1 right parenthesis , so 2 superscript 2 k baseline minus 1 is not prime.

22k−1=(2k+1)(2k−1), so 22k−1 is prime.
2 superscript 2 k baseline minus 1 equals left parenthesis 2 superscript k baseline plus 1 right parenthesis left parenthesis 2 superscript k baseline minus 1 right parenthesis , so 2 superscript 2 k baseline minus 1 is prime.

22k−1=(22+1)(2k−1), so 22k−1 is prime.

Answer 1

To determine which response properly uses a polynomial identity to detect if \( 2^{2k} - 1 \) is prime, we can analyze the given statements.

First, let's note that:

1. **Polynomial identity involving \( 2^k \)**:
- The identity \( 2^{2k} - 1 = (2^k - 1)(2^k + 1) \) can be used to factor \( 2^{2k} - 1 \) whenever \( k > 0 \).
2. If \( k \geq 1 \), both \( 2^k - 1 \) and \( 2^k + 1 \) are greater than 1, implying that \( 2^{2k} - 1 \) is not prime.

Now, let's consider the responses:

1. **First Response**:
- " \( 2^{2k-1}=(2^2+1)(2^{k-1}) \), so \( 2^{2k-1} \) is not prime."
- This is incorrect because \( 2^{2k-1} \) does not equal \( (2^2 + 1)(2^{k-1}) \).

2. **Second Response**:
- " \( 2^{2k-1}=(2^{k}+1)(2^{k}-1) \), so \( 2^{2k-1} \) is not prime."
- This is also incorrect because \( 2^{2k-1} \neq (2^k + 1)(2^k - 1) \).

3. **Third Response**:
- " \( 2^{2k-1}=(2^k+1)(2^k-1) \), so \( 2^{2k-1} \) is prime."
- This is misleading; since \( k \geq 1 \), \( 2^k - 1 \) and \( 2^k + 1 \) are both greater than 1, making this statement inaccurate.

4. **Fourth Response**:
- " \( 2^{2k-1}=(2^2+1)(2^{k-1}) \), so \( 2^{2k-1} \) is prime."
- This is also incorrect because the identity is wrong and does not lead to a valid conclusion about primality.

Given the options, none of the responses provide a correct application of a polynomial identity to definitively determine if \( 2^{2k-1} \) is prime or not.

However, the closest to being correct is the second response, as it attempts a polynomial factorization (albeit incorrectly).

To clarify, the correct way to show \( 2^{2k} - 1 \) is not prime for \( k \geq 1 \) using polynomial identity is:
- \( 2^{2k} - 1 = (2^k - 1)(2^k + 1) \), which shows \( 2^{2k} - 1 \) cannot be prime since both factors are greater than 1 for \( k \geq 1 \), but this is not explicitly stated in any of the responses.

In conclusion, none of the given responses correctly and clearly conclude whether \( 2^{2k-1} \) is prime or not using a polynomial identity.