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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

$$\mathrm{a},\mathrm{b},\mathrm{c}\:\in\mathbb{N}\:\mathrm{such}\:\mathrm{that}\:\frac{\mathrm{a}\sqrt{\mathrm{3}}\:+\mathrm{b}}{\mathrm{b}\sqrt{\mathrm{3}}+\mathrm{c}}\:\in\:\mathrm{Q},\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:\in\:\mathbb{Z} \\ $$

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)

$$\mathrm{Fact}\::\:\sqrt{\mathrm{3}}\:\notin\mathrm{Q}\:,\:\mathrm{p}+\mathrm{q}\sqrt{\mathrm{3}}\:\in\mathrm{Q}\:\mathrm{if}\:\mathrm{q}=\mathrm{0} \\ $$$$\mathrm{consider}\:\frac{\mathrm{a}\sqrt{\mathrm{3}}+\mathrm{b}}{\mathrm{b}\sqrt{\mathrm{3}}+\mathrm{c}}\:\in\mathrm{Q}\:\Rightarrow\frac{\left(\mathrm{a}\sqrt{\mathrm{3}}+\mathrm{b}\right)\left(\mathrm{b}\sqrt{\mathrm{3}}−\mathrm{c}\right)}{\mathrm{3b}−\mathrm{c}} \\ $$$$=\frac{\mathrm{3ab}−\mathrm{bc}+\sqrt{\mathrm{3}}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{ac}\right)}{\mathrm{3b}−\mathrm{c}}\:\in\mathrm{Q}\: \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\::\:\mathrm{b}^{\mathrm{2}} −\mathrm{ac}\:=\:\mathrm{0}\:\mathrm{or}\:\mathrm{b}^{\mathrm{2}} =\mathrm{ac} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:= \\ $$$$\frac{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\right)}{\mathrm{a}+\mathrm{b}+\mathrm{c}}= \\ $$$$\frac{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}+\mathrm{b}^{\mathrm{2}} +\mathrm{bc}\right)}{\mathrm{a}+\mathrm{b}+\mathrm{c}}= \\ $$$$\frac{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{2b}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:= \\ $$$$\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)−\mathrm{2b}\:=\:\mathrm{a}−\mathrm{b}+\mathrm{c}\:\in\:\mathbb{Z} \\ $$$$\left(\mathrm{proved}\right) \\ $$

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

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

$$\mathrm{But}\:\mathrm{b}^{\mathrm{2}} \geqslant\mathrm{2ac},\:\mathrm{you}\:\mathrm{only}\:\mathrm{considered} \\ $$$$\mathrm{b}^{\mathrm{2}} =\mathrm{2ac} \\ $$

Commented by bobhans last updated on 13/Sep/20

you are wrong

$$\mathrm{you}\:\mathrm{are}\:\mathrm{wrong} \\ $$

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