Prove-that-5-divide-n-4n-2-1-6n-2-1-for-any-natural-number-n-

Question Number 158240 by HongKing last updated on 01/Nov/21

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{5}\:\mathrm{divide} \\ $$$$\mathrm{n}\left(\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{6n}^{\mathrm{2}} \:+\:\mathrm{1}\right) \\ $$$$\mathrm{for}\:\mathrm{any}\:\mathrm{natural}\:\mathrm{number}\:\boldsymbol{\mathrm{n}} \\ $$

Answered by Rasheed.Sindhi last updated on 01/Nov/21

A(n)=n(4n^2 + 1)(6n^2 + 1) n is possibly equal to one of the following: 5m,5m+1,5m+2,5m+3,5m+4 n=5m: A(5m)=5m(4(5m)^2 +1)(6(5m)^2 +1) Obviously 5 ∣ A(5m) n=5m+1: For n=5m+1 The factor 4n^2 + 1 of A(n) = 4(5m+1)^2 +1=100m^2 +40m+5 =5(20m^2 +8m+1) ∴ 5 ∣ A(5m+1) n=5m+2: For n=5m+2 the factor 6n^2 +1 =6(5m+2)^2 +1=150m^2 +120m+25 =5(30m^2 +24m+5) ∴ 5∣A(5m+2) n=5m+3: The factor 6n^2 +1 of A(n) =6(5m+3)^2 +1=150m^2 +180m+55 =5(30m^2 +36m+11) ∴ 5 ∣ A(5m+3) n=5m+4: The factor 4n^2 +1 of A(n) =4(5m+4)^2 +1=100m^2 +160+65 =5(20m^2 +32m+13) ∴ 5 ∣ A(5m+4) HENCE 5 ∣ A(n) in all possible cases

$$\mathrm{A}\left(\mathrm{n}\right)=\mathrm{n}\left(\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{6n}^{\mathrm{2}} \:+\:\mathrm{1}\right) \\ $$$$\mathrm{n}\:{is}\:{possibly}\:{equal}\:{to}\:{one}\:{of}\:{the}\:{following}: \\ $$$$\:\mathrm{5m},\mathrm{5m}+\mathrm{1},\mathrm{5m}+\mathrm{2},\mathrm{5m}+\mathrm{3},\mathrm{5m}+\mathrm{4} \\ $$$$\mathrm{n}=\mathrm{5m}: \\ $$$$\mathrm{A}\left(\mathrm{5m}\right)=\mathrm{5m}\left(\mathrm{4}\left(\mathrm{5m}\right)^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{6}\left(\mathrm{5m}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${Obviously}\:\:\mathrm{5}\:\mid\:\mathrm{A}\left(\mathrm{5m}\right)\: \\ $$$$\mathrm{n}=\mathrm{5m}+\mathrm{1}: \\ $$$${For}\:{n}=\mathrm{5m}+\mathrm{1} \\ $$$$\mathcal{T}{he}\:{factor}\:\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{1}\:{of}\:\mathrm{A}\left(\mathrm{n}\right) \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{4}\left(\mathrm{5m}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{100m}^{\mathrm{2}} +\mathrm{40m}+\mathrm{5} \\ $$$$\:\:\:\:\:\:\:=\mathrm{5}\left(\mathrm{20m}^{\mathrm{2}} +\mathrm{8m}+\mathrm{1}\right) \\ $$$$\therefore\:\mathrm{5}\:\mid\:\mathrm{A}\left(\mathrm{5m}+\mathrm{1}\right) \\ $$$$\mathrm{n}=\mathrm{5m}+\mathrm{2}: \\ $$$${For}\:\mathrm{n}=\mathrm{5m}+\mathrm{2}\:{the}\:{factor}\:\mathrm{6n}^{\mathrm{2}} +\mathrm{1} \\ $$$$=\mathrm{6}\left(\mathrm{5m}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{150m}^{\mathrm{2}} +\mathrm{120m}+\mathrm{25} \\ $$$$=\mathrm{5}\left(\mathrm{30m}^{\mathrm{2}} +\mathrm{24m}+\mathrm{5}\right) \\ $$$$\therefore\:\mathrm{5}\mid\mathrm{A}\left(\mathrm{5m}+\mathrm{2}\right) \\ $$$$\mathrm{n}=\mathrm{5m}+\mathrm{3}: \\ $$$$\mathcal{T}{he}\:{factor}\:\mathrm{6n}^{\mathrm{2}} +\mathrm{1}\:{of}\:\mathrm{A}\left(\mathrm{n}\right) \\ $$$$=\mathrm{6}\left(\mathrm{5m}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{150m}^{\mathrm{2}} +\mathrm{180m}+\mathrm{55} \\ $$$$=\mathrm{5}\left(\mathrm{30m}^{\mathrm{2}} +\mathrm{36m}+\mathrm{11}\right) \\ $$$$\therefore\:\mathrm{5}\:\mid\:\mathrm{A}\left(\mathrm{5m}+\mathrm{3}\right) \\ $$$$\mathrm{n}=\mathrm{5m}+\mathrm{4}: \\ $$$$\:\mathcal{T}{he}\:{factor}\:\mathrm{4n}^{\mathrm{2}} +\mathrm{1}\:{of}\:\mathrm{A}\left(\mathrm{n}\right) \\ $$$$=\mathrm{4}\left(\mathrm{5m}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{100m}^{\mathrm{2}} +\mathrm{160}+\mathrm{65} \\ $$$$=\mathrm{5}\left(\mathrm{20m}^{\mathrm{2}} +\mathrm{32m}+\mathrm{13}\right) \\ $$$$\therefore\:\mathrm{5}\:\mid\:\mathrm{A}\left(\mathrm{5m}+\mathrm{4}\right) \\ $$$$ \\ $$$$\mathcal{HENCE} \\ $$$$\mathrm{5}\:\mid\:\mathrm{A}\left(\mathrm{n}\right)\:\mathrm{in}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{cases} \\ $$

Commented by HongKing last updated on 01/Nov/21

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{cool} \\ $$

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