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.

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

Commented by1549442205 last updated on 01/Jul/20

Thank you sir a lot.You are welcome.

Commented by1549442205 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 εA is evenε true for ∀n=2k  From i)and ii) follows A is even ∀n  Sir,is it all right?Thank you sir!

Commented byMJS last updated on 30/Jun/20

looks right to 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)