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Question Number 90661 by Cynosure last updated on 25/Apr/20

show that (n^4 −n^2 ) is divisible by 12

$${show}\:{that}\:\left({n}^{\mathrm{4}} −{n}^{\mathrm{2}} \right)\:{is}\:{divisible}\:{by}\:\mathrm{12} \\ $$

Answered by MJS last updated on 25/Apr/20

n^4 −n^2 =n^2 (n^2 −1)=n^2 (n−1)(n+1)=f(n) (1) n=6k f(n)=36k^2 (6k−1)(6k+1)=12×3k^2 (6k−1)(6k+1) (2) n=6k+1 f(n)=(6k+1)^2 (6k)(6k+2)=12(6k+1)^2 k(3k+1) (3) n=6k+2 f(n)=(6k+2)^2 (6k+1)(6k+3)=12(3k+1)^2 (6k+1)(2k+1) (4) n=6k+3 f(n)=(6k+3)^2 (6k+2)(6k+4)=12(2k+1)(6k+3)(3k+1)(3k+2) (5) n=6k+4 f(n)=(6k+4)^2 (6k+3)(6k+5)=12(3k+2)^2 (2k+1)(6k+5) (6) n=6k+5 f(n)=(6k+5)^2 (6k+4)(6k+6)=12(6k+5)^2 (3k+2)(k+1)

$${n}^{\mathrm{4}} −{n}^{\mathrm{2}} ={n}^{\mathrm{2}} \left({n}^{\mathrm{2}} −\mathrm{1}\right)={n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)={f}\left({n}\right) \\ $$$$\left(\mathrm{1}\right)\:{n}=\mathrm{6}{k} \\ $$$${f}\left({n}\right)=\mathrm{36}{k}^{\mathrm{2}} \left(\mathrm{6}{k}−\mathrm{1}\right)\left(\mathrm{6}{k}+\mathrm{1}\right)=\mathrm{12}×\mathrm{3}{k}^{\mathrm{2}} \left(\mathrm{6}{k}−\mathrm{1}\right)\left(\mathrm{6}{k}+\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:{n}=\mathrm{6}{k}+\mathrm{1} \\ $$$${f}\left({n}\right)=\left(\mathrm{6}{k}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{6}{k}\right)\left(\mathrm{6}{k}+\mathrm{2}\right)=\mathrm{12}\left(\mathrm{6}{k}+\mathrm{1}\right)^{\mathrm{2}} {k}\left(\mathrm{3}{k}+\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:{n}=\mathrm{6}{k}+\mathrm{2} \\ $$$${f}\left({n}\right)=\left(\mathrm{6}{k}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{6}{k}+\mathrm{1}\right)\left(\mathrm{6}{k}+\mathrm{3}\right)=\mathrm{12}\left(\mathrm{3}{k}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{6}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$\left(\mathrm{4}\right)\:{n}=\mathrm{6}{k}+\mathrm{3} \\ $$$${f}\left({n}\right)=\left(\mathrm{6}{k}+\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{6}{k}+\mathrm{2}\right)\left(\mathrm{6}{k}+\mathrm{4}\right)=\mathrm{12}\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{6}{k}+\mathrm{3}\right)\left(\mathrm{3}{k}+\mathrm{1}\right)\left(\mathrm{3}{k}+\mathrm{2}\right) \\ $$$$\left(\mathrm{5}\right)\:{n}=\mathrm{6}{k}+\mathrm{4} \\ $$$${f}\left({n}\right)=\left(\mathrm{6}{k}+\mathrm{4}\right)^{\mathrm{2}} \left(\mathrm{6}{k}+\mathrm{3}\right)\left(\mathrm{6}{k}+\mathrm{5}\right)=\mathrm{12}\left(\mathrm{3}{k}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{6}{k}+\mathrm{5}\right) \\ $$$$\left(\mathrm{6}\right)\:{n}=\mathrm{6}{k}+\mathrm{5} \\ $$$${f}\left({n}\right)=\left(\mathrm{6}{k}+\mathrm{5}\right)^{\mathrm{2}} \left(\mathrm{6}{k}+\mathrm{4}\right)\left(\mathrm{6}{k}+\mathrm{6}\right)=\mathrm{12}\left(\mathrm{6}{k}+\mathrm{5}\right)^{\mathrm{2}} \left(\mathrm{3}{k}+\mathrm{2}\right)\left({k}+\mathrm{1}\right) \\ $$

Answered by JDamian last updated on 25/Apr/20

m=n^4 −n^2 =n^2 (n^2 −1)=(n−1)n^2 (n+1) m is the product of three sequential natural numbers. Therefore: 1 m=3k 2.1 when n is even ⇒ m=4k 2.2 when n is odd ⇒(n−1) and (n+1) are both even ⇒ m=4k 3 From above, m=12k

$${m}={n}^{\mathrm{4}} −{n}^{\mathrm{2}} ={n}^{\mathrm{2}} \left({n}^{\mathrm{2}} −\mathrm{1}\right)=\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} \left({n}+\mathrm{1}\right) \\ $$$$ \\ $$$$\boldsymbol{{m}}\:{is}\:{the}\:{product}\:{of}\:{three}\:{sequential} \\ $$$${natural}\:{numbers}.\:{Therefore}: \\ $$$$\mathrm{1}\:\:\:\:\:\:{m}=\mathrm{3}{k} \\ $$$$\mathrm{2}.\mathrm{1}\:{when}\:\boldsymbol{{n}}\:{is}\:{even}\:\Rightarrow\:{m}=\mathrm{4}{k} \\ $$$$\mathrm{2}.\mathrm{2}\:{when}\:\boldsymbol{{n}}\:{is}\:{odd}\:\Rightarrow\left({n}−\mathrm{1}\right)\:{and}\:\left({n}+\mathrm{1}\right)\:{are} \\ $$$$\:\:\:\:\:\:\:\:{both}\:{even}\:\Rightarrow\:{m}=\mathrm{4}{k} \\ $$$$\mathrm{3}\:\:\:\:\:\:{From}\:{above},\:{m}=\mathrm{12}{k} \\ $$

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