for any positive integer n, prove that n^3-n is divisible by 6
Answers
Step-by-step explanation:
Check whether n
3
−n is divisible by 3.
n
3
−n=n(n+1)(n−1)
When a number is divided by 3 then by the remainder theorem, the remainder obtained is either 0 or 1 or 2.
n=3p or n=3p+1 or n=3p+2, where p is some integer.
If n=3p, then the number is divisible by 3.
If n=3p+1, then n−1=3p+1−1=3p. The number is divisible by 3.
If n=3p+2, then n+1=3p+2+1=3(p+1). The number is divisible by 3.
So, any number in the form of n
3
−n=n(n+1)(n−1) is divisible by 3.
Check whether n
3
−n is divisible by 2.
When a number is divided by 2, the remainder obtained is either 0 or 1 by the remainder theorem.
n=2p or n=2p+1, where p is some integer.
If n=2p, then the number is divisible by 2.
If n=2p+1 then n−1=2p+1−1=2p. The number is divisible by 2.
So, any number in the form of n
3
−n=n(n+1)(n−1) is divisible by 2.
Since, the given number n
3
−n=n(n+1)(n−1) is divisible by both 3 and 2. Therefore, according to the divisibility rule of 6, the given number is divisible by 6.
Hence, n
3
−n=n(n+1)(n−1) is divisible by 6.
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