Alright, let's work through this problem step-by-step.
We are trying to find the greatest number that always divides the product of the predecessor and successor of an odd natural number greater than 1.
1. Identify the odd natural number:
Let [tex]\( n \)[/tex] be an odd natural number such that [tex]\( n > 1 \)[/tex].
2. Define the predecessor and successor:
The predecessor of [tex]\( n \)[/tex] is [tex]\( n - 1 \)[/tex] and the successor of [tex]\( n \)[/tex] is [tex]\( n + 1 \)[/tex].
3. Formulate the product of predecessor and successor:
The product of the predecessor and successor is given by [tex]\( (n - 1)(n + 1) \)[/tex].
4. Expand the product:
[tex]\[
(n - 1)(n + 1) = n^2 - 1
\][/tex]
This is a difference of squares.
5. Analyze [tex]\( n \)[/tex]:
Since [tex]\( n \)[/tex] is odd, we can represent [tex]\( n \)[/tex] as [tex]\( 2k + 1 \)[/tex], where [tex]\( k \)[/tex] is an integer.
6. Substitute [tex]\( n = 2k + 1 \)[/tex] into the product:
[tex]\[
(2k + 1)^2 - 1
\][/tex]
Expanding this gives:
[tex]\[
(2k + 1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k = 4k(k + 1)
\][/tex]
7. Examine the expression:
The expression [tex]\( 4k(k + 1) \)[/tex] is the product of 4 and two consecutive integers, [tex]\( k \)[/tex] and [tex]\( k + 1 \)[/tex].
8. Determine the divisibility:
- The factor [tex]\( 4k \)[/tex] indicates the product includes [tex]\( 4 \)[/tex].
- Since [tex]\( k \)[/tex] and [tex]\( k + 1 \)[/tex] are consecutive integers, one of them is always even, meaning [tex]\( k(k + 1) \)[/tex] includes at least a factor of 2.
So, combining these, the entire product [tex]\( 4k(k + 1) \)[/tex] includes the factors 4 and 2 from the product of two consecutive numbers. This means:
[tex]\[
4 \times 2 = 8
\][/tex]
Hence, the greatest number that always divides [tex]\( n^2 - 1 \)[/tex], where [tex]\( n \)[/tex] is any odd natural number greater than 1, is:
[tex]\[
\boxed{8}
\][/tex]
