Divisible by 3?

Multiplying and dividing by 2, we get:

$\dfrac{2n(2n-1)(2n+1)}{2}$

The product of 3 consecutive positive integers is always divisible by 3 and $2n(2n-1)(2n+1)$ is even and hence is also divisible by 6.

Alternatively, we can generalise that the product of $n$ consecutive positive integers is always divisible by $n!$ .

So $2n(2n-1)(2n+1)$ is divisible by $3!$ which is $6$ .

$\displaystyle \Rightarrow 2n(2n-1)(2n+1) \equiv 0 \pmod{6}$

$\displaystyle \Rightarrow n(2n-1)(2n+1) \equiv 0 \pmod{3}$

$n \equiv 0,1,2 (\mod 3)$

Case 1: If $n \equiv 0 (\mod 3)$

Trivial case, The number will be divisible for all positive $n$

Case 2: $n \equiv 1 (\mod 3)$

Then $2n-1 \equiv 1+1-1 \equiv 1(\mod 3)$

and $2n+1 \equiv 1+1+1 \equiv 0 (\mod 3)$

Thus the number is divisible by 3 for $n \equiv 1 (\mod 3)$

Case 3: $n \equiv 2 (\mod 3)$

Then, $2n-1 \equiv 2+2-1 \equiv 3 \equiv 0 (\mod 3)$

Thus the number is always divisible by 3 for $n \equiv 2 (\mod 3)$

We have checked all cases and we conclude $n(2n+1)(2n-1)$ is always divisible by 3 for all positive n.

$n$ can be of form $3n$ , $3n + 1$ or $3n+2$ . On substituting these into the expression, all yield a multiple of 3.

Let's use some Induction here:

CASE I $(n = 1):$ $(1)(2(1)-1)(2(1)+1) = (1)(1)(3) = 3 \Rightarrow \boxed{TRUE}.$

CASE II $(n = k):$ $(k)(2k-1)(2k+1) = 4k^3 - k = 3N \Rightarrow$ assume to be TRUE.

CASE III $(n = k+1):$ $4k^3 - k = 3N;$

or $4[(k+1)-1]^3 - [(k+1)-1] = 3N;$

or $4[(k+1)^3 - 3(k+1)^2 (1) + 3(k+1)(1)^2 - 1] - (k+1) + 1 = 3N;$

or $4(k+1)^3 - (k+1) - 3 - 3[4(k+1)^2] + 3[4(k+1)] = 3N;$

or $4(k+1)^3 - (k+1) = 3[N + 1 + 4(k+1)^3 - 4(k+1)] \Rightarrow \boxed{TRUE}.$

Hence, $3|n(2n-1)(2n+1)$ for all $n \in \mathbb{N}.$

$\mathbb{Q.} \mathbb{E.} \mathbb{D.}$

One can also prove this by Induction.