show that ∀ n ∈ N, 12 divise n^2 (n^4 −1)


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Question Number 128012 by mathocean1 last updated on 03/Jan/21

$s h o w t h a t \forall n \in N, 12 d i v i s e n^{2} (n^{4} - 1)$
	Answered by MJS_new last updated on 03/Jan/21

	$n = 2 k \Rightarrow 2 ∣ n \Rightarrow 4 ∣ n^{2}$ $n = 2 k + 1 \Rightarrow n^{4} - 1 = 8 k (k - 1) (2 k^{2} - 2 k + 1) \Rightarrow$ $\Rightarrow 4 ∣ n^{4} - 1$ $\Rightarrow 4 ∣ n^{2} (n^{4} - 1)$ $n = 3 k \Rightarrow 3 ∣ n^{2}$ $n = 3 k + 1 \Rightarrow n^{4} - 1 = 3 k (3 k + 2) (9 k^{2} + 6 k + 2) \Rightarrow$ $\Rightarrow 3 ∣ n^{4} - 1$ $n = 3 k + 2 \Rightarrow n^{4} - 1 = 3 k (k + 1) (3 k + 1) (9 k^{2} + 12 k + 5) \Rightarrow$ $\Rightarrow 3 ∣ n^{4} - 1$ $\Rightarrow 3 ∣ n^{2} (n^{4} - 1)$ $\Rightarrow 12 ∣ n^{2} (n^{4} - 1)$
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