For any positive integer n,

Solution:

To Prove: For any positive integer n, n3 − n is divisible by 6. 

Proof: Let n be any positive integer.

$\Rightarrow n^{3}-n=(n-1)(n)(n+1)$

Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5

 

If n = 6q

Then, $(n-1) n(n+1)=(6 q-1) 6 q(6 q+1)$

which is divisble by 6

If n = 6q + 1

Then, $(n-1) n(n+1)=(6 q)(6 q+1)(6 q+2)$

$\Rightarrow$ which is divisble by 6

If n = 6q + 2

Then, $(n-1) n(n+1)=(6 q+1)(6 q+2)(6 q+3)$

$\Rightarrow(n-1) n(n+1)=6(6 q+1)(3 q+1)(2 q+1)$

which is divisble by 6

Similarly we can prove others.

 

Hence it is proved that for any positive integer n, n3 − n is divisible by 6.