If a is an odd integer prove that a^2 − 1 is always divisible by 8.
Solution
Proof.
If a is divided by 4 then the possible set of remainders are 0, 1, 2 and 3.
Since a is odd so the remainder when divided by 4 cannot be 0 or 2.
So if a is an odd integer then the remainder when divided by 4 is 1 or 3.
Now we do case analysis:
Case 1: Let the remainder be 1.
So a = 4k + 1 for some integer k.
Thus a2 = (4k + 1)2 = 16k2 + 8k + 1 = 8(2k2 + k) + 1 So a2 − 1 = 8(2k2 + k) and so a2 − 1 is divisible by 8
Case 2: Let the remainder be 3.
So a = 4k + 3 for some integer k.
Thus a2 = (4k + 3)2 = 16k2 + 24k + 9 = 8(2k2 + 3k + 1) + 1 So a2 − 1 = 8(2k2 + 3k + 1) and so a2 − 1 is divisible by 8
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