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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,c∈Nsuchthata3+bb3+c∈Q,showthata2+b2+c2a+b+c∈Z

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:3∉Q,p+q3∈Qifq=0considera3+bb3+c∈Q⇒(a3+b)(b3−c)3b−c=3ab−bc+3(b2−ac)3b−c∈Qitshouldbe:b2−ac=0orb2=acnowwehavea2+b2+c2a+b+c=(a+b+c)2−2(ab+ac+bc)a+b+c=(a+b+c)2−2(ab+b2+bc)a+b+c=(a+b+c)2−2b(a+b+c)a+b+c=(a+b+c)−2b=a−b+c∈Z(proved)

Commented by Aina Samuel Temidayo last updated on 12/Sep/20

But b^2 ≥2ac, you only considered b^2 =2ac

Butb2⩾2ac,youonlyconsideredb2=2ac

Commented by bobhans last updated on 13/Sep/20

you are wrong

youarewrong

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