By factoring the expression, check that is always divisible by 6 for all natural numbers . Give reasons.

Question

Homi · Accepted Answer

Understand the Question

A number is divisible by $6$ if and only if it is divisible by both $2$ and $3$ (since $\gcd(2, 3) = 1$ ).
Factoring the expression $n^3 - n$ yields $(n - 1)n(n + 1)$ , which represents the product of three consecutive integers.
Among any three consecutive integers:
- At least one must be a multiple of $2$ (even).
- Exactly one must be a multiple of $3$ .
Therefore, their product is always divisible by $2 \times 3 = 6$ .

Step 1 · Factorise the Expression

Given expression: $n^3 - n$

Taking $n$ as a common factor: $n^3 - n = n(n^2 - 1)$

Using the algebraic identity $a^2 - b^2 = (a - 1)(a + 1)$ : $n^3 - n = (n - 1)n(n + 1)$

Here, $(n - 1)$ , $n$ , and $(n + 1)$ are three consecutive integers.

Step 2 · Check Divisibility by 2 and 3

For any three consecutive natural numbers $(n - 1)$ , $n$ , and $(n + 1)$ :

Divisibility by 2: Out of any two consecutive numbers, one is always even. Hence, at least one of $(n - 1), n, (n + 1)$ is divisible by $2$ . $\implies (n - 1)n(n + 1) \text{ is divisible by } 2$
Divisibility by 3: Out of any three consecutive numbers, exactly one must be a multiple of $3$ . $\implies (n - 1)n(n + 1) \text{ is divisible by } 3$

Since $2$ and $3$ are coprime (i.e., $\gcd(2, 3) = 1$ ): $(n - 1)n(n + 1) \text{ is divisible by } 2 \times 3 = 6$

Answer

Hence, $n^3 - n = (n - 1)n(n + 1)$ is always divisible by $6$ for all natural numbers $n$ .

Common Mistakes

Case for $n = 1$ : For $n = 1$ , $1^3 - 1 = 0$ . Remember that $0$ is divisible by every non-zero integer, including $6$ ( $0 = 6 \times 0$ ).
Incomplete Divisibility Check: Showing divisibility only by $2$ or only by $3$ is not enough; you must show divisibility by both coprime factors to conclude divisibility by $6$ .

Question 12