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Mod 2: If N is odd, it is of the form, (2k + 1) for some k. (2k+1)² = 4k² + 4k + 1. Subtract the 1, and you have 4k² + 4k. 4k² + 4k = 4 (k² + k) Regardless of whether k is odd or even, k² is the same, so k² + k is even. Let k² + k = 2 j. We have 4 * 2 * j, which is divisible by 8. |
Mod 4: And odd number is either 4k + 1 or 4k + 3 for some k (4k + 1)² = 16k² + 8k + 1 Subtract the one and the rest is divisible by 8 (4k + 3)² = 16k² + 24k + 9 Subtract the 1 16k + 24k + 8 which is divisible by 8. |
Mathematical induction: Assume (2k + 1)² - 1 is divisible by 8. The next odd number is (2k + 3). (2k + 3)² - (2k+1)² = (4k² + 12k + 9) - (4k² + 4k + 1) = 8k + 8. So if (2k+1)² - 1 is divisible by 8, so is (2k+3)² - 1 because they differ by 8k + 8 which is divisible by 8. For the base case: 1² = 1, 1² - 1 = 0 is divisible by 8, or if you prefer, start with 3: 3² - 1 = 9 - 1 = 8. Therefore it is true for all odd numbers. |
