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Question Number 128012 by mathocean1 last updated on 03/Jan/21
show that ∀ n ∈ N, 12 divise n^2 (n^4 −1)
$${show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N},\:\mathrm{12}\:{divise}\:{n}^{\mathrm{2}} \left({n}^{\mathrm{4}} −\mathrm{1}\right) \\ $$
Answered by MJS_new last updated on 03/Jan/21
n=2k ⇒ 2∣n ⇒ 4∣n^2 n=2k+1 ⇒ n^4 −1=8k(k−1)(2k^2 −2k+1) ⇒ ⇒ 4∣n^4 −1 ⇒ 4∣n^2 (n^4 −1) n=3k ⇒ 3∣n^2 n=3k+1 ⇒ n^4 −1=3k(3k+2)(9k^2 +6k+2) ⇒ ⇒ 3∣n^4 −1 n=3k+2 ⇒ n^4 −1=3k(k+1)(3k+1)(9k^2 +12k+5) ⇒ ⇒ 3∣n^4 −1 ⇒ 3∣n^2 (n^4 −1) ⇒ 12∣n^2 (n^4 −1)
$${n}=\mathrm{2}{k}\:\Rightarrow\:\mathrm{2}\mid{n}\:\Rightarrow\:\mathrm{4}\mid{n}^{\mathrm{2}} \\ $$$${n}=\mathrm{2}{k}+\mathrm{1}\:\Rightarrow\:{n}^{\mathrm{4}} −\mathrm{1}=\mathrm{8}{k}\left({k}−\mathrm{1}\right)\left(\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{4}\mid{n}^{\mathrm{4}} −\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{4}\mid{n}^{\mathrm{2}} \left({n}^{\mathrm{4}} −\mathrm{1}\right) \\ $$$$ \\ $$$${n}=\mathrm{3}{k}\:\Rightarrow\:\mathrm{3}\mid{n}^{\mathrm{2}} \\ $$$${n}=\mathrm{3}{k}+\mathrm{1}\:\Rightarrow\:{n}^{\mathrm{4}} −\mathrm{1}=\mathrm{3}{k}\left(\mathrm{3}{k}+\mathrm{2}\right)\left(\mathrm{9}{k}^{\mathrm{2}} +\mathrm{6}{k}+\mathrm{2}\right)\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{3}\mid{n}^{\mathrm{4}} −\mathrm{1} \\ $$$${n}=\mathrm{3}{k}+\mathrm{2}\:\Rightarrow\:{n}^{\mathrm{4}} −\mathrm{1}=\mathrm{3}{k}\left({k}+\mathrm{1}\right)\left(\mathrm{3}{k}+\mathrm{1}\right)\left(\mathrm{9}{k}^{\mathrm{2}} +\mathrm{12}{k}+\mathrm{5}\right)\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{3}\mid{n}^{\mathrm{4}} −\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{3}\mid{n}^{\mathrm{2}} \left({n}^{\mathrm{4}} −\mathrm{1}\right) \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{12}\mid{n}^{\mathrm{2}} \left({n}^{\mathrm{4}} −\mathrm{1}\right) \\ $$

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