If n = 4k + 3, does 8 divide n2 - 1? (Assume that n and k represent integers.) O Yes, because n2 - 1 = (4k + 3)2 - 1 = (16k? + 9) – 1 = 16k? + 8 = 8(2k? + 1), and 2k² + 1 is an integer because sums and products of integers are integers. O Yes, because 4k + 3 is an odd integer and 8 is a divisor of 4k + 3. O Yes, because n2 - 1 = (4k + 3)2 - 1 = (16k2 + 24k + 9) – 1 = 16k? + 24k + 8 = 8(2k2 + 3k + 1), and 2k2 + 3k + 1 is an integer because sums and products of integers are integers. O Yes, because 8 is even, 4k + 3 is odd, the square of any odd integer is odd, and an odd number minus 1 is even. O No, because even though n2 - 1 = (16k2 + 9) -1 = 16k2 + 8 is a sum of even numbers and hence even, there are even integers that are not divisible by 8. O No, because even though n2 - 1 = (4k + 3)2 - 1 = (16k? + 24k + 9) – 1 = 16k? + 24k + 8 is a sum of even numbers and hence even, there are even integers that are not divisible by 8.

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A: As per Bartleby guidelines we can solve any one question at a time. Hope you understand.

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