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:
- 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 \( k > 0 \).
- 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:
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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 \( (2^2 + 1)(2^{k-1}) \).
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Second Response:
- " \( 2^{2k-1}=(2^{k}+1)(2^{k}-1) \), so \( 2^{2k-1} \) is not prime."
- This is also incorrect because \( 2^{2k-1} \neq (2^k + 1)(2^k - 1) \).
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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.
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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.