8/(n^{2}-1)∀n∈odd integers
Was the question asking to prove by induction that n^{2}-1 is divisible by 8 for odd positive n?
To use induction, we first shall prove the base case: n=1.
n=1 satisfies the desired result since (1)^{2}-1=0, which is divisible by 8.
Second, we appeal to the inductive hypothesis: Assume that for n odd, n^{2}-1 is divisible by 8. Then, we wish to show that (n+2)^{2}-1 is also divisible by 8 (where n+2 is the odd number immediately after n).
We can show the above as follows: (n+2)^{2}-1=n^{2}+4n+4-1=n^{2}-1+4(n+1). Now, n^{2}-1 is divisible by 8 by the inductive hypothesis. Moreover, n+1 is even (since n is odd by assumption) and since any even multiple of 4 is divisible by 8, 4(n+1) must be divisible by 8.
The above statement is obvious when you re-express n+1 as 2N for some other integer N (since any even number can be expressed thus). Then 4(n+1)=4*2N=8N, which is clearly divisible by 8 (8N/8=N).
Finally, the sum of two individual numbers divisible by 8 makes the sum divisible by 8, making (n+2)^{2}-1 divisible by 8.
Having shown the base case, and having shown that given an odd number satisfying the property, the odd number immediately following it will also satisfy it, we have proven the desired result.
Q.E.D.
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