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Question Number 2748 by Rasheed Soomro last updated on 26/Nov/15
Show that ((3^(2n+1) +5^(2n+1) )/8) is an integer for n∈Z^+ .
$${Show}\:{that}\:\frac{\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{5}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{8}}\:{is}\:{an}\:{integer}\:{for}\:{n}\in\mathbb{Z}^{+} \:. \\ $$
Answered by prakash jain last updated on 26/Nov/15
f(n)=3^(2n+1) +5^(2n+1) f(1)=27+125=152=19×8 Let us say f(n) 3^(2n+1) +5^(2n+1) is divisible by 8. f(n+1)=3^(2n+3) +5^(2n+3) f(n+1)−f(n)=3^(2n+3) −3^(2n+1) +5^(2n+3) −5^(2n+1) =3^(2n+1) (8)+5^(2n+1) (24) =8(3^(2n+1) +3∙5^(2n+1) ) ∵ ((f(n))/8)=k⇒f(n)=8k, k∈N f(n+1)=8(3^(2n+1) +3∙5^(2n+1) )+8k f(n+1)=8(3^(2n+1) +3∙5^(2n+1) +k) ∴ f(n+1) is divisible by 8 or ((f(n+1))/8) is an integer.
$${f}\left({n}\right)=\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{5}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{27}+\mathrm{125}=\mathrm{152}=\mathrm{19}×\mathrm{8} \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:{f}\left({n}\right)\:\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{5}^{\mathrm{2}{n}+\mathrm{1}} \:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{8}. \\ $$$${f}\left({n}+\mathrm{1}\right)=\mathrm{3}^{\mathrm{2}{n}+\mathrm{3}} +\mathrm{5}^{\mathrm{2}{n}+\mathrm{3}} \\ $$$${f}\left(\mathrm{n}+\mathrm{1}\right)−{f}\left({n}\right)=\mathrm{3}^{\mathrm{2}{n}+\mathrm{3}} −\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{5}^{\mathrm{2}{n}+\mathrm{3}} −\mathrm{5}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{8}\right)+\mathrm{5}^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{24}\right) \\ $$$$=\mathrm{8}\left(\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}\centerdot\mathrm{5}^{\mathrm{2n}+\mathrm{1}} \right) \\ $$$$\because\:\frac{{f}\left({n}\right)}{\mathrm{8}}={k}\Rightarrow{f}\left({n}\right)=\mathrm{8}{k},\:{k}\in\mathbb{N} \\ $$$${f}\left({n}+\mathrm{1}\right)=\mathrm{8}\left(\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}\centerdot\mathrm{5}^{\mathrm{2}{n}+\mathrm{1}} \right)+\mathrm{8}{k} \\ $$$${f}\left({n}+\mathrm{1}\right)=\mathrm{8}\left(\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}\centerdot\mathrm{5}^{\mathrm{2}{n}+\mathrm{1}} +{k}\right) \\ $$$$\therefore\:{f}\left({n}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{8}\:\mathrm{or} \\ $$$$\frac{{f}\left({n}+\mathrm{1}\right)}{\mathrm{8}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}. \\ $$
Commented by RasheedAhmad last updated on 26/Nov/15
Nice!
$$\mathcal{N}{ice}! \\ $$

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