If n is an odd integer, then show that \(n^2 - 1\) is divisible by 8.
Divisible by 8.
Step 1: Express an odd integer.
Any odd integer can be written as
\(n = 2k + 1\), where \(k\) is an integer.
Step 2: Expand \(n^2 - 1\).
\(n^2 - 1 = (2k + 1)^2 - 1\)
\(= 4k^2 + 4k + 1 - 1\)
\(= 4k^2 + 4k\)
\(= 4k(k + 1)\).
Step 3: Analyze the product \(k(k+1)\).
The numbers \(k\) and \(k+1\) are consecutive integers.
In any two consecutive integers, one is always even.
Therefore, their product \(k(k+1)\) is always even.
Step 4: Multiply by 4.
Since \(k(k+1)\) is even, we can write
\(k(k+1) = 2m\), for some integer \(m\).
So, \(n^2 - 1 = 4 \times 2m = 8m\).
Conclusion.
This shows that \(n^2 - 1\) is a multiple of 8. Hence, it is divisible by 8.