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Question Number 78948 by TawaTawa last updated on 21/Jan/20

Prove by mathematical induction that. n^4 + 4n^2 + 11 is divisible by 16

$$\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}. \\ $$$$\:\:\:\mathrm{n}^{\mathrm{4}} \:+\:\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{11}\:\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{16} \\ $$

Commented by john santu last updated on 21/Jan/20

let p(n) = n^4 +4n^2 +11 for n=1 ⇒1+4+11=16 ∣16 (true) suppose n=k such that p(k)=k^4 +4k^2 +11 = u(mod 16) we can proof that for n=k+1 divisible by 16 k^4 +4k^2 +11+(k+1)^4 +4(k+1)+11 next..

$${let}\:{p}\left({n}\right)\:=\:{n}^{\mathrm{4}} +\mathrm{4}{n}^{\mathrm{2}} +\mathrm{11} \\ $$$${for}\:{n}=\mathrm{1}\:\Rightarrow\mathrm{1}+\mathrm{4}+\mathrm{11}=\mathrm{16}\:\mid\mathrm{16}\:\left({true}\right) \\ $$$${suppose}\:{n}={k}\:{such}\:{that}\: \\ $$$${p}\left({k}\right)={k}^{\mathrm{4}} +\mathrm{4}{k}^{\mathrm{2}} +\mathrm{11}\:=\:{u}\left({mod}\:\mathrm{16}\right) \\ $$$${we}\:{can}\:{proof}\:{that}\:{for}\: \\ $$$${n}={k}+\mathrm{1}\:{divisible}\:{by}\:\mathrm{16} \\ $$$${k}^{\mathrm{4}} +\mathrm{4}{k}^{\mathrm{2}} +\mathrm{11}+\left({k}+\mathrm{1}\right)^{\mathrm{4}} +\mathrm{4}\left({k}+\mathrm{1}\right)+\mathrm{11} \\ $$$${next}.. \\ $$

Commented by TawaTawa last updated on 21/Jan/20

It is for k + 1 i don′t get sir

$$\mathrm{It}\:\mathrm{is}\:\mathrm{for}\:\:\mathrm{k}\:+\:\mathrm{1}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{sir} \\ $$

Commented by mind is power last updated on 21/Jan/20

error in this n=2k didn′t worck ever

$${error}\:{in}\:{this} \\ $$$${n}=\mathrm{2}{k}\:{didn}'{t}\:{worck}\:{ever} \\ $$$$ \\ $$

Commented by TawaTawa last updated on 21/Jan/20

That means the question is wrong sir?

$$\mathrm{That}\:\mathrm{means}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{sir}? \\ $$

Commented by mind is power last updated on 21/Jan/20

yeah

$${yeah} \\ $$

Commented by john santu last updated on 21/Jan/20

and ...in fact not proved miss the equation error. k^4 +4k^2 +11 not divisible by 16. i.q let k=2 ⇒16+16+11≠∣16

$${and}\:\:\:\:...{in}\:{fact}\:{not}\:{proved}\:{miss} \\ $$$${the}\:{equation}\:{error}.\:{k}^{\mathrm{4}} +\mathrm{4}{k}^{\mathrm{2}} +\mathrm{11} \\ $$$${not}\:{divisible}\:{by}\:\mathrm{16}. \\ $$$${i}.{q}\:{let}\:{k}=\mathrm{2}\:\Rightarrow\mathrm{16}+\mathrm{16}+\mathrm{11}\neq\mid\mathrm{16} \\ $$

Commented by TawaTawa last updated on 21/Jan/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mind is power last updated on 21/Jan/20

first Methode n=2k ⇒n^4 +4n^2 +11=16k^4 +16k^2 +11≡11(16) error sir

$${first}\:{Methode} \\ $$$${n}=\mathrm{2}{k} \\ $$$$\Rightarrow{n}^{\mathrm{4}} +\mathrm{4}{n}^{\mathrm{2}} +\mathrm{11}=\mathrm{16}{k}^{\mathrm{4}} +\mathrm{16}{k}^{\mathrm{2}} +\mathrm{11}\equiv\mathrm{11}\left(\mathrm{16}\right) \\ $$$${error}\:{sir}\: \\ $$

Commented by TawaTawa last updated on 21/Jan/20

Sir what of this n^4 + 4n^2 + 11 is a multiple of 16 for all odd positive integral

$$\mathrm{Sir}\:\mathrm{what}\:\mathrm{of}\:\mathrm{this} \\ $$$$\:\:\mathrm{n}^{\mathrm{4}} \:+\:\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{11}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\:\mathrm{16}\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{odd}\:\mathrm{positive}\:\mathrm{integral} \\ $$

Commented by TawaTawa last updated on 21/Jan/20

Help me prove this

$$\mathrm{Help}\:\mathrm{me}\:\mathrm{prove}\:\mathrm{this} \\ $$

Commented by mind is power last updated on 21/Jan/20

ifn=(2k+1) n^4 =(2k+1)^4 =16k^4 +32k^3 +24k^2 +8k+1 n^2 =4k^2 +4k+1 n^4 +4n^2 +1=16k^4 +32k^3 +24k^2 +8k+1+4(4k^2 +4k+1)+11 =16k^4 +32k^3 +40k^2 +24k+16 =16(k^4 +2k^3 )+16+8k(5k+3)=8k(5k+3)mod(16) k(5k+3)=k(5k+5−2)=5k(k+1)−2k=2m ⇒8k(5k+3)=8.2m=16m=0(16) ⇒16∣n^4 +4n^2 +11⇒n=(2k+1)is True

$${ifn}=\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$${n}^{\mathrm{4}} =\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{4}} =\mathrm{16}{k}^{\mathrm{4}} +\mathrm{32}{k}^{\mathrm{3}} +\mathrm{24}{k}^{\mathrm{2}} +\mathrm{8}{k}+\mathrm{1} \\ $$$${n}^{\mathrm{2}} =\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1} \\ $$$${n}^{\mathrm{4}} +\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}=\mathrm{16}{k}^{\mathrm{4}} +\mathrm{32}{k}^{\mathrm{3}} +\mathrm{24}{k}^{\mathrm{2}} +\mathrm{8}{k}+\mathrm{1}+\mathrm{4}\left(\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}\right)+\mathrm{11} \\ $$$$=\mathrm{16}{k}^{\mathrm{4}} +\mathrm{32}{k}^{\mathrm{3}} +\mathrm{40}{k}^{\mathrm{2}} +\mathrm{24}{k}+\mathrm{16} \\ $$$$=\mathrm{16}\left({k}^{\mathrm{4}} +\mathrm{2}{k}^{\mathrm{3}} \right)+\mathrm{16}+\mathrm{8}{k}\left(\mathrm{5}{k}+\mathrm{3}\right)=\mathrm{8}{k}\left(\mathrm{5}{k}+\mathrm{3}\right){mod}\left(\mathrm{16}\right) \\ $$$${k}\left(\mathrm{5}{k}+\mathrm{3}\right)={k}\left(\mathrm{5}{k}+\mathrm{5}−\mathrm{2}\right)=\mathrm{5}{k}\left({k}+\mathrm{1}\right)−\mathrm{2}{k}=\mathrm{2}{m} \\ $$$$\Rightarrow\mathrm{8}{k}\left(\mathrm{5}{k}+\mathrm{3}\right)=\mathrm{8}.\mathrm{2}{m}=\mathrm{16}{m}=\mathrm{0}\left(\mathrm{16}\right) \\ $$$$\Rightarrow\mathrm{16}\mid{n}^{\mathrm{4}} +\mathrm{4}{n}^{\mathrm{2}} +\mathrm{11}\Rightarrow{n}=\left(\mathrm{2}{k}+\mathrm{1}\right){is}\:{True} \\ $$$$ \\ $$

Commented by TawaTawa last updated on 22/Jan/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mind is power last updated on 22/Jan/20

y′re Welcom Sir Withe pleasur

$${y}'{re}\:{Welcom}\:{Sir}\:{Withe}\:{pleasur} \\ $$

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