Which of the following properly uses a polynomial identity to detect if 22k−1 is prime?(1 point)Responses22k−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 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=(22+1)(2k−1), so 22k−1 is prime.

Question

Which of the following properly uses a polynomial identity to detect if  22k−1  is prime?(1 point)Responses22k−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 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=(22+1)(2k−1), so 22k−1 is prime.

GPT-4o mini · Answer

To determine if $ 2^{2k} - 1 $ is prime using a polynomial identity, we can use the identity: $$ a^2 - b^2 = (a - b)(a + b) $$ In our case, we can let $ a = 2^k $ and $ b = 1 $. Therefore: $$ 2^{2k} - 1 = (2^k - 1)(2^k + 1) $$ This shows that $ 2^{2k} - 1 $ can be factored into two non-trivial factors: $ (2^k - 1) $ and $ (2^k + 1) $. For $ k \geq 1 $, both factors are greater than 1, which means $ 2^{2k} - 1 $ is not prime. Based on the provided responses, the correct response is: $$ 2^{2k} - 1 = (2^k + 1)(2^k - 1), \text{ so } 2^{2k} - 1 \text{ is not prime.} $$ Therefore, the correct option is: $$ 2^{2k} - 1 = (2^k + 1)(2^k - 1), \text{ so } 2^{2k} - 1 \text{ is not prime.} $$