Question

For any positive integer n, prove that n3-n is divisible by 3.

Answer 1

Not possible because if n=2 then n3-n=4 which is not divisible by 3.

Answer 2

Question:

For any positive integer n, prove that  n³ - n  is divisible by 3.

Proof by mathematical induction:

1. Let n = 1.

$n^3-n=1^3-1=1-1=0$

As 0 is divisible by all integers, it is also divisible by 3.

2. Let n = 2.

$n^2-n=2^3-2=8-2=6$

6 is divisible by 3.

3. Let n = k.

Assume that  k³ - k  is divisible by 3.

4. Let n = k + 1.

$\Longrightarrow\ n^3-n \\ \\ \Longrightarrow\ (k+1)^3-(k+1) \\ \\ \Longrightarrow\ k^3+3k^2+3k+1-k-1 \\ \\ \Longrightarrow\ k^3-k+3k^2+3k \\ \\ \Longrightarrow\ k^3-k+3(k^2+k)$

Consider the last step. It is assumed earlier that  k³ - k  is a multiple of 3. To this,  3(k² + k), which is also multiple of 3, is added.

Thus,  n³ - n  s divisible by 3, for any positive integer n.

Hence proved!!!