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Question Number 101052 by 1549442205 last updated on 30/Jun/20

Given a=^3 (√(7+(√(50)))),b=^3 (√(7−(√(50)))).Prove  that a^7 +b^7 is an even number.

$$\mathrm{Given}\:\mathrm{a}=\:^{\mathrm{3}} \sqrt{\mathrm{7}+\sqrt{\mathrm{50}}},\mathrm{b}=\:^{\mathrm{3}} \sqrt{\mathrm{7}−\sqrt{\mathrm{50}}}.\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} \mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{number}. \\ $$

Answered by MJS last updated on 30/Jun/20

a=((7+5(√2)))^(1/3) =1+(√2)  b=((7−5(√2)))^(1/3) =1−(√2)  we can show that (p+q(√r))^n +(p−q(√r))^n   with p, q∈Z and r, n∈N must always be  an even number  α=p; β=q(√r)  (α+β)^n +(α−β)^n  leads to cancelling out  some terms and doubling the remaining  terms ⇒ (α+β)^n +(α−β)^n =2k  I haven′t got the time to type the proof but  it′s obvious

$${a}=\sqrt[{\mathrm{3}}]{\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${b}=\sqrt[{\mathrm{3}}]{\mathrm{7}−\mathrm{5}\sqrt{\mathrm{2}}}=\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:\mathrm{that}\:\left({p}+{q}\sqrt{{r}}\right)^{{n}} +\left({p}−{q}\sqrt{{r}}\right)^{{n}} \\ $$$$\mathrm{with}\:{p},\:{q}\in\mathbb{Z}\:\mathrm{and}\:{r},\:{n}\in\mathbb{N}\:\mathrm{must}\:\mathrm{always}\:\mathrm{be} \\ $$$$\mathrm{an}\:\mathrm{even}\:\mathrm{number} \\ $$$$\alpha={p};\:\beta={q}\sqrt{{r}} \\ $$$$\left(\alpha+\beta\right)^{{n}} +\left(\alpha−\beta\right)^{{n}} \:\mathrm{leads}\:\mathrm{to}\:\mathrm{cancelling}\:\mathrm{out} \\ $$$$\mathrm{some}\:\mathrm{terms}\:\mathrm{and}\:\mathrm{doubling}\:\mathrm{the}\:\mathrm{remaining} \\ $$$$\mathrm{terms}\:\Rightarrow\:\left(\alpha+\beta\right)^{{n}} +\left(\alpha−\beta\right)^{{n}} =\mathrm{2}{k} \\ $$$$\mathrm{I}\:\mathrm{haven}'\mathrm{t}\:\mathrm{got}\:\mathrm{the}\:\mathrm{time}\:\mathrm{to}\:\mathrm{type}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{but} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{obvious} \\ $$

Commented by 1549442205 last updated on 01/Jul/20

Thank you sir a lot.You are welcome.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{a}\:\mathrm{lot}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

Commented by 1549442205 last updated on 30/Jun/20

Thank you sir a lot .Please,I get permission  to submit a way as follows:  i)If n is odd then A=(p+q(√r))^(2k+1) +(p−q(√r))^(2k+1)   =2p[(p+q(√r))^(2k) −(p+q)^(2k−1) (p−q(√r))+  ...(p−q(√r))^(2k) ]⇒A is even  ii)If n =2k then A=[(p+q(√r))^k +(p−q(√r))^k ]^2   +for k=1⇒A=2p^2 +2q^2 r⇒A is even  +Suppose A is even ∀n≤2k i.e   A_n =(p+q(√r))^n +(p−q(√r))^n is even ∀n≤2k  Then A_(2n+2) =[(p+q(√r))^(k+1) +(p−q(√r))^(k+1) ]^2   −2(p^2 −q^2 r)^(k+1) =(2M)^2 −2(p^2 −q^2 r)^(k+1) =2Q  which shows that ”A is even” true for ∀n=2k  From i)and ii) follows A is even ∀n  Sir,is it all right?Thank you sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{a}\:\mathrm{lot}\:.\mathrm{Please},\mathrm{I}\:\mathrm{get}\:\mathrm{permission} \\ $$$$\mathrm{to}\:\mathrm{submit}\:\mathrm{a}\:\mathrm{way}\:\mathrm{as}\:\mathrm{follows}: \\ $$$$\left.\mathrm{i}\right)\mathrm{If}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{then}\:\mathrm{A}=\left(\mathrm{p}+\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{2k}+\mathrm{1}} +\left(\mathrm{p}−\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{2k}+\mathrm{1}} \\ $$$$=\mathrm{2p}\left[\left(\mathrm{p}+\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{2k}} −\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2k}−\mathrm{1}} \left(\mathrm{p}−\mathrm{q}\sqrt{\mathrm{r}}\right)+\right. \\ $$$$\left....\left(\mathrm{p}−\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{2k}} \right]\Rightarrow\mathrm{A}\:\mathrm{is}\:\mathrm{even} \\ $$$$\left.\mathrm{ii}\right)\mathrm{If}\:\mathrm{n}\:=\mathrm{2k}\:\mathrm{then}\:\mathrm{A}=\left[\left(\mathrm{p}+\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{k}} +\left(\mathrm{p}−\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{k}} \right]^{\mathrm{2}} \\ $$$$+\mathrm{for}\:\mathrm{k}=\mathrm{1}\Rightarrow\mathrm{A}=\mathrm{2p}^{\mathrm{2}} +\mathrm{2q}^{\mathrm{2}} \mathrm{r}\Rightarrow\mathrm{A}\:\mathrm{is}\:\mathrm{even} \\ $$$$+\mathrm{Suppose}\:\mathrm{A}\:\mathrm{is}\:\mathrm{even}\:\forall\mathrm{n}\leqslant\mathrm{2k}\:\mathrm{i}.\mathrm{e}\: \\ $$$$\mathrm{A}_{\mathrm{n}} =\left(\mathrm{p}+\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{n}} +\left(\mathrm{p}−\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{n}} \mathrm{is}\:\mathrm{even}\:\forall\mathrm{n}\leqslant\mathrm{2k} \\ $$$$\mathrm{Then}\:\mathrm{A}_{\mathrm{2n}+\mathrm{2}} =\left[\left(\mathrm{p}+\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{k}+\mathrm{1}} +\left(\mathrm{p}−\mathrm{q}\sqrt{\mathrm{r}}\right)^{\mathrm{k}+\mathrm{1}} \right]^{\mathrm{2}} \\ $$$$−\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} −\mathrm{q}^{\mathrm{2}} \mathrm{r}\right)^{\mathrm{k}+\mathrm{1}} =\left(\mathrm{2M}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} −\mathrm{q}^{\mathrm{2}} \mathrm{r}\right)^{\mathrm{k}+\mathrm{1}} =\mathrm{2Q} \\ $$$$\mathrm{which}\:\mathrm{shows}\:\mathrm{that}\:''\mathrm{A}\:\mathrm{is}\:\mathrm{even}''\:\mathrm{true}\:\mathrm{for}\:\forall\mathrm{n}=\mathrm{2k} \\ $$$$\left.\mathrm{F}\left.\mathrm{rom}\:\mathrm{i}\right)\mathrm{and}\:\mathrm{ii}\right)\:\mathrm{follows}\:\mathrm{A}\:\mathrm{is}\:\mathrm{even}\:\forall\mathrm{n} \\ $$$$\mathrm{Sir},\mathrm{is}\:\mathrm{it}\:\mathrm{all}\:\mathrm{right}?\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$

Commented by MJS last updated on 30/Jun/20

looks right to me

$$\mathrm{looks}\:\mathrm{right}\:\mathrm{to}\:\mathrm{me} \\ $$

Answered by 1549442205 last updated on 03/Jul/20

ab=^3 (√((7+(√(50)))(7−(√(50)))))=−1.Hence,  a^3 +b^3 =14=(a+b)^3 −3ab(a+b)  =(a+b)^3 +3(a+b)⇒(a+b)^3 +3(a+b)−14=0  ⇔(a+b−2)[(a+b)^2 +2(a+b)+7]=0   ⇒a+b=2.From  a^7 +b^7 =(a+b)(a^6 −a^5 b+a^4 b^2 −a^3 b^3 +a^2 b^4 −ab^5 +b^6 )  =2(a^6 −a^5 b+a^4 b^2 −a^3 b^3 +a^2 b^4 −ab^5 +b^6 )  which shows that a^7 +b^7 is odd number(q.e.d)  other way:  a^4 +b^4 =(a^2 +b^2 )^2 −2a^2 b^2 =[(a+b)^2 −2ab]^2 −2a^2 b^2   =(2^2 −2(−1))^2 −2×1=6^2 −2=34  a^7 +b^7 =(a^3 +b^3 )(a^4 +b^4 )−a^3 b^3 (a+b)  =14×34−(−1)^3 ×2=478,so a^7 +b^7   is a evev number(q.e.d)

$$\mathrm{ab}=\:^{\mathrm{3}} \sqrt{\left(\mathrm{7}+\sqrt{\mathrm{50}}\right)\left(\mathrm{7}−\sqrt{\mathrm{50}}\right)}=−\mathrm{1}.\mathrm{Hence}, \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{14}=\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$=\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{a}+\mathrm{b}\right)\Rightarrow\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{a}+\mathrm{b}\right)−\mathrm{14}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{a}+\mathrm{b}−\mathrm{2}\right)\left[\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)+\mathrm{7}\right]=\mathrm{0} \\ $$$$\:\Rightarrow\mathrm{a}+\mathrm{b}=\mathrm{2}.\mathrm{From} \\ $$$$\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} =\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{6}} −\mathrm{a}^{\mathrm{5}} \mathrm{b}+\mathrm{a}^{\mathrm{4}} \mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{4}} −\mathrm{ab}^{\mathrm{5}} +\mathrm{b}^{\mathrm{6}} \right) \\ $$$$=\mathrm{2}\left(\mathrm{a}^{\mathrm{6}} −\mathrm{a}^{\mathrm{5}} \mathrm{b}+\mathrm{a}^{\mathrm{4}} \mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{4}} −\mathrm{ab}^{\mathrm{5}} +\mathrm{b}^{\mathrm{6}} \right) \\ $$$$\mathrm{which}\:\mathrm{shows}\:\mathrm{that}\:\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} \mathrm{is}\:\mathrm{odd}\:\mathrm{number}\left(\mathrm{q}.\mathrm{e}.\mathrm{d}\right) \\ $$$$\boldsymbol{\mathrm{other}}\:\boldsymbol{\mathrm{way}}: \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{b}}^{\mathrm{4}} =\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} =\left[\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{ab}}\right]^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} \\ $$$$=\left(\mathrm{2}^{\mathrm{2}} −\mathrm{2}\left(−\mathrm{1}\right)\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{1}=\mathrm{6}^{\mathrm{2}} −\mathrm{2}=\mathrm{34} \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{b}}^{\mathrm{7}} =\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} \right)\left(\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{b}}^{\mathrm{4}} \right)−\boldsymbol{\mathrm{a}}^{\mathrm{3}} \boldsymbol{\mathrm{b}}^{\mathrm{3}} \left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right) \\ $$$$=\mathrm{14}×\mathrm{34}−\left(−\mathrm{1}\right)^{\mathrm{3}} ×\mathrm{2}=\mathrm{478},\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{b}}^{\mathrm{7}} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{evev}}\:\boldsymbol{\mathrm{number}}\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

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