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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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