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a-b-c-N-such-that-a-3-b-b-3-c-Q-show-that-a-2-b-2-c-2-a-b-c-Z-




Question Number 113187 by bobhans last updated on 11/Sep/20
a,b,c ∈N such that ((a(√3) +b)/(b(√3)+c)) ∈ Q, show  that ((a^2 +b^2 +c^2 )/(a+b+c)) ∈ Z
a,b,cNsuchthata3+bb3+cQ,showthata2+b2+c2a+b+cZ
Answered by bemath last updated on 11/Sep/20
Fact : (√3) ∉Q , p+q(√3) ∈Q if q=0  consider ((a(√3)+b)/(b(√3)+c)) ∈Q ⇒(((a(√3)+b)(b(√3)−c))/(3b−c))  =((3ab−bc+(√3)(b^2 −ac))/(3b−c)) ∈Q   it should be : b^2 −ac = 0 or b^2 =ac  now we have ((a^2 +b^2 +c^2 )/(a+b+c)) =  (((a+b+c)^2 −2(ab+ac+bc))/(a+b+c))=  (((a+b+c)^2 −2(ab+b^2 +bc))/(a+b+c))=  (((a+b+c)^2 −2b(a+b+c))/(a+b+c)) =  (a+b+c)−2b = a−b+c ∈ Z  (proved)
Fact:3Q,p+q3Qifq=0considera3+bb3+cQ(a3+b)(b3c)3bc=3abbc+3(b2ac)3bcQitshouldbe:b2ac=0orb2=acnowwehavea2+b2+c2a+b+c=(a+b+c)22(ab+ac+bc)a+b+c=(a+b+c)22(ab+b2+bc)a+b+c=(a+b+c)22b(a+b+c)a+b+c=(a+b+c)2b=ab+cZ(proved)
Commented by Aina Samuel Temidayo last updated on 12/Sep/20
But b^2 ≥2ac, you only considered  b^2 =2ac
Butb22ac,youonlyconsideredb2=2ac
Commented by bobhans last updated on 13/Sep/20
you are wrong
youarewrong

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