Final answer:
To prove that 100 consecutive integers are not perfect squares, we use a direct proof by showing that the gaps between perfect squares become larger as numbers increase, easily allowing for a span of 100 non-perfect squares.
Step-by-step explanation:
The student asked to prove that there are 100 consecutive integers that are not perfect squares. This can be proven using a direct proof. We know that perfect squares increase as the squares of consecutive positive integers. For example, the perfect squares of integers 1, 2, 3, and so on are 1, 4, 9, 16, etc. The difference between consecutive perfect squares increases as we move up the number line, meaning the gap between them gets larger.
Let's take 32² = 1024 and 33² = 1089. The difference between these two perfect squares is 65, and no integers between 1024 and 1089 are perfect squares. Therefore, there exists at least one sequence of 65 consecutive numbers that are not perfect squares. Notably, the difference between 99² (9801) and 100² (10000) is 199. Thereby, between 9801 and 10000, there are 199 consecutive integers that are not perfect squares, which is well over the 100 consecutive integers required.
Therefore, by choosing an integer such that the difference between its square and the square of the next integer is greater than 100, we can ensure that there is a sequence of 100 consecutive integers that are not perfect squares. This fulfills the requirement, and the answer is a direct proof, denoted as option (C).