Source: Mathcenter Contest / Oly - Thai Forum 2008 R1 p4 https://artofproblemsolving.com/community/c3196914_mathcenter_contest
Tags: number theory, algebra, rational, Integer
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Let $a,b$ and $c$ be positive integers that $$\frac{a\sqrt{3}+b}{b\sqrt3+c}$$is a rational number, show that $$\frac{a^2+b^2+c^2}{a+b+ c}$$is an integer.
(Anonymous314)
Rationalizing the quotient we get
$$\frac{(a\sqrt3 + b)(b\sqrt3 - c)}{3b^2 - c^2} = \frac{3ab - bc + (b^2-ac)\sqrt3}{3b^2-c^2} \in\mathbb{Q}.$$Because $\sqrt3$ is irrational, we have $b^2 = ca$, but now:
$$\frac{a^2+b^2+c^2}{a+b+c} = \frac{(a+b+c)^2 - 2(ab+bc+ca)}{a+b+c} = a+b+c - 2\frac{ab+bc+b^2}{a+b+c} = a+ b+ c - 2b \in \mathbb{Z}.$$
Finnish Mathematics Competition 2004, Final Round, Problem 2