Show [n(n^2−1)]/3 is an integer

Question

Use the division algorithm to show [n(n^2−1)]/3 is an integer, for all n ∈ Z.

Answer 1

 

Division algorithm used in inside the proof by induction:

 

1*2*3 = 6 which is divisible by 3
2*3*4 = 24 which is divisible by 3
3*4*5 = 60 which is divisible by 3

The given (induction hypothesis) guarantees the following statement holds:
For some positive integer k, k(K+1)(k+2) is divisible by 3.

Prove: (k+1)(k+2)(k+3) is divisible by 3.

Proof:

Given by induction hypothesis, k(k+1)(k+2) is divisible by 3.

By division algorithm:
k(k+1)(k+2) = 3 * N for some integer N,and there is no remainder

(k+1)(k+2) = 3 * N / k <--- divides both sides by positive integer k

Note that by closure property of integer addition and multiplication,
the left side is an integer. So 3*N/k MUST be an integer.

(k+1)(k+2)(k+3) = 3 * N/k * (k+3) is also an integer and a multiple of 3.

[end of proof]

Answer 2

n(n²-1) = n(n+1)(n-1) which is the product of three consecutive integers. 

Exactly one of any three consecutive integers is divisible by three, therefore the expression [n(n^2−1)]/3 must be an integer when n is an integer.