In this problem, we prove a basic divisibility property of integers. We are required to show that for every positive integer n, the expression n³ − n is always divisible by 6.
Question
For any positive integer n, prove that n³ − n is divisible by 6.
Solution
Consider the expression
n³ − n.
Taking n common, we get
n³ − n = n(n² − 1).
Now, n² − 1 can be written as
n² − 1 = (n − 1)(n + 1).
Therefore,
n³ − n = n(n − 1)(n + 1).
Here, n − 1, n, and n + 1 are three consecutive integers.
Among any three consecutive integers, one integer is divisible by 3.
Also, among any two consecutive integers, one must be even.
Hence, the product n(n − 1)(n + 1) is divisible by both 2 and 3.
Since 6 = 2 × 3, the expression n³ − n is divisible by 6.
Conclusion
Therefore, for any positive integer n, the value of n³ − n is always divisible by 6.
Hence proved.