Answer:
We can prove this statement using direct proof.
Assume that n is an odd integer. Then, we can express n as 2k + 1, where k is an integer.
Now, we want to show that 8∣(n^2 - 1), which means that there exists an integer m such that n^2 - 1 = 8m.
Substituting n = 2k + 1, we get:
n^2 - 1 = (2k + 1)^2 - 1
Simplifying this expression, we get:
n^2 - 1 = 4k^2 + 4k
Now, we can factor out a 4 from the right-hand side of the equation:
n^2 - 1 = 4(k^2 + k)
We can further factor the expression k^2 + k as:
k^2 + k = k(k+1)
So, we have:
n^2 - 1 = 4k(k+1)
Now, we want to show that 8 divides n^2 - 1, which means that we need to show that 2^3 divides 4k(k+1).
Since k and k+1 are two consecutive integers, one of them must be even. Therefore, we can write k(k+1) as the product of an even integer and an odd integer.
Case 1: k is even
If k is even, then k+1 is odd. Therefore, k(k+1) is even and can be written as 2j, where j is an integer.
Substituting this expression into n^2 - 1 = 4k(k+1), we get:
n^2 - 1 = 4(2j)
n^2 - 1 = 8j
Therefore, 8 divides n^2 - 1.
Case 2: k is odd
If k is odd, then k+1 is even. Therefore, k(k+1) is even and can be written as 2j, where j is an integer.
Substituting this expression into n^2 - 1 = 4k(k+1), we get:
n^2 - 1 = 4(2j)
n^2 - 1 = 8j
Therefore, 8 divides n^2 - 1.
Since we have shown that n^2 - 1 is divisible by 8 in both cases, we can conclude that for every odd integer n, 8 divides n^2 - 1.
Thus, we have proven the statement.
Step-by-step explanation:
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