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Question Number 108453 by gospelkenny last updated on 17/Aug/20

If  R=(7+4(√3))^(2n)  = I+f, where I ∈ N  and   0<f<1, then R(1−f) equals

$$\mathrm{If}\:\:{R}=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} \:=\:{I}+{f},\:\mathrm{where}\:{I}\:\in\:{N} \\ $$ $$\mathrm{and}\:\:\:\mathrm{0}<{f}<\mathrm{1},\:\mathrm{then}\:{R}\left(\mathrm{1}−{f}\right)\:\mathrm{equals} \\ $$

Answered by 1549442205PVT last updated on 17/Aug/20

Set Q=(7−4(√3))^(2n) ⇒(√(RQ))=[(7+4(√3))^n (7−4(√3))^n =1(∗)  We prove that  S= (√R)+(√Q)∈N^∗ .Indeed,  i)For n=1we get S=(7+4(√3))+(7−4(√3))  =14⇒State is true for n=1  ii)Suppose State is true ∀ n=1...k^(−)    ⇒S_k =(7+4(√3))^k +(7−4(√3))^k =m_k ∈N  iii)Consider S_(k+1) =(7+4(√3))^(k+1) +(7−4(√3))^(k+1)   =[(7+4(√3))^k +(7−4(√3))^k ][(7+4(√3))+(7−4(√3))]  −[(7+4(√3))×(7−4(√3))][(7+4(√3))^(k−1) +(7−4(√3))^(k−1) ]  =14S_k −S_(k−1) =14m_k −m_(k−1) ∈N^∗   (by the introduction hypothesis m_k ,m_(k−1) ∈N^∗ )  This shows that the state is also true for n=k+1  Hence by the imtroduction principle  the state is true for ∀n∈N^∗ which   means ((√R)+(√Q))∈N^∗ ⇒((√R)+(√Q))^2 ∈N^∗   ⇔R+Q=((√R)+(√Q))^2 −2∈N^∗   ⇒(7+4(√3))^(2n) +(7−4(√3))^(2n) =q∈N^∗  (∗∗)  On the other hands,by above we have  [(7+4(√3))^(2n) ×(7−4(√3))^(2n) ]=1  ⇒0<(7−4(√3))^(2n) =(1/((7+4(√3))^(2n) ))<1  Therefore ,if R=(7+4(√3))^(2n) =I+f and  I∈N and 0<f<1 then by (∗∗)  I=q and f=(7−4(√3))^(2n)  .It follows that  R(1−f)=R−Rf=(7+4(√3))^(2n) −[(7+4(√3))^(2n) ×(7−4(√3))^(2n) ]  =(7+4(√3))^(2n) −1   Thus,R(1−f)=(7+4(√3))^(2n) −1

$$\mathrm{Set}\:\mathrm{Q}=\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2n}} \Rightarrow\sqrt{\mathrm{RQ}}=\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{n}} \left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{n}} =\mathrm{1}\left(\ast\right)\right. \\ $$ $$\mathrm{We}\:\mathrm{prove}\:\mathrm{that}\:\:\mathrm{S}=\:\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}}\in\mathbb{N}^{\ast} .\mathrm{Indeed}, \\ $$ $$\left.\mathrm{i}\right)\mathrm{For}\:\mathrm{n}=\mathrm{1we}\:\mathrm{get}\:\mathrm{S}=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)+\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$ $$=\mathrm{14}\Rightarrow\mathrm{State}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{1} \\ $$ $$\left.\mathrm{ii}\right)\mathrm{Suppose}\:\mathrm{State}\:\mathrm{is}\:\mathrm{true}\:\forall\:\mathrm{n}=\overline {\mathrm{1}...\mathrm{k}} \\ $$ $$\:\Rightarrow\mathrm{S}_{\mathrm{k}} =\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}} +\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}} =\mathrm{m}_{\mathrm{k}} \in\mathbb{N} \\ $$ $$\left.\mathrm{iii}\right)\mathrm{Consider}\:\mathrm{S}_{\mathrm{k}+\mathrm{1}} =\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}+\mathrm{1}} +\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}+\mathrm{1}} \\ $$ $$=\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}} +\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}} \right]\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)+\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)\right] \\ $$ $$−\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)×\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)\right]\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}−\mathrm{1}} +\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{k}−\mathrm{1}} \right] \\ $$ $$=\mathrm{14S}_{\mathrm{k}} −\mathrm{S}_{\mathrm{k}−\mathrm{1}} =\mathrm{14m}_{\mathrm{k}} −\mathrm{m}_{\mathrm{k}−\mathrm{1}} \in\mathbb{N}^{\ast} \\ $$ $$\left(\mathrm{by}\:\mathrm{the}\:\mathrm{introduction}\:\mathrm{hypothesis}\:\mathrm{m}_{\mathrm{k}} ,\mathrm{m}_{\mathrm{k}−\mathrm{1}} \in\mathbb{N}^{\ast} \right) \\ $$ $$\mathrm{This}\:\mathrm{shows}\:\mathrm{that}\:\mathrm{the}\:\mathrm{state}\:\mathrm{is}\:\mathrm{also}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}+\mathrm{1} \\ $$ $$\mathrm{Hence}\:\mathrm{by}\:\mathrm{the}\:\mathrm{imtroduction}\:\mathrm{principle} \\ $$ $$\mathrm{the}\:\mathrm{state}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\forall\mathrm{n}\in\mathbb{N}^{\ast} \mathrm{which}\: \\ $$ $$\mathrm{means}\:\left(\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}}\right)\in\mathbb{N}^{\ast} \Rightarrow\left(\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}}\right)^{\mathrm{2}} \in\mathbb{N}^{\ast} \\ $$ $$\Leftrightarrow\mathrm{R}+\mathrm{Q}=\left(\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}}\right)^{\mathrm{2}} −\mathrm{2}\in\mathbb{N}^{\ast} \\ $$ $$\Rightarrow\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}\boldsymbol{{n}}} +\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}\boldsymbol{{n}}} =\boldsymbol{\mathrm{q}}\in\mathbb{N}^{\ast} \:\left(\ast\ast\right) \\ $$ $$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands},\mathrm{by}\:\mathrm{above}\:\mathrm{we}\:\mathrm{have} \\ $$ $$\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} ×\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} \right]=\mathrm{1} \\ $$ $$\Rightarrow\mathrm{0}<\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} =\frac{\mathrm{1}}{\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} }<\mathrm{1} \\ $$ $$\mathrm{Therefore}\:,\mathrm{if}\:\mathrm{R}=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} =\mathrm{I}+\mathrm{f}\:\mathrm{and} \\ $$ $$\mathrm{I}\in\mathbb{N}\:\mathrm{and}\:\mathrm{0}<\mathrm{f}<\mathrm{1}\:\mathrm{then}\:\mathrm{by}\:\left(\ast\ast\right) \\ $$ $$\boldsymbol{\mathrm{I}}=\boldsymbol{\mathrm{q}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{f}}=\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}\boldsymbol{{n}}} \:.\mathrm{It}\:\mathrm{follows}\:\mathrm{that} \\ $$ $$\mathrm{R}\left(\mathrm{1}−\mathrm{f}\right)=\mathrm{R}−\mathrm{Rf}=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} −\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} ×\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} \right] \\ $$ $$=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}\boldsymbol{{n}}} −\mathrm{1}\: \\ $$ $$\mathrm{Thus},\boldsymbol{\mathrm{R}}\left(\mathrm{1}−\boldsymbol{\mathrm{f}}\right)=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}\boldsymbol{{n}}} −\mathrm{1} \\ $$

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