MHB Is the Expression an Integer When Rational Conditions Apply?

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The discussion revolves around a problem involving positive integers a, b, and c, where the expression (a√3 + b) / (b√3 + c) is rational. Participants are tasked with proving that (a² + b² + c²) / (a + b + c) is an integer under these conditions. Members Opalg and kaliprasad provided correct solutions to the problem. The focus is on the mathematical reasoning and proofs related to the rationality of the expression. The thread emphasizes the significance of understanding rational conditions in integer expressions.
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Here is this week's POTW:

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Given that $a,\,b,\,c$ are positive integers such that $\dfrac{a\sqrt{3}+b}{b\sqrt{3}+c}$ is a rational number.

Show that $\dfrac{a^2+b^2+c^2}{a+b+c}$ is an integer.

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Congratulations to the following members for their correct solution!(Cool)

1. Opalg
2. kaliprasad

Solution from Opalg:
If $\dfrac{a\sqrt3 + b}{b\sqrt3 + c} = r$ (rational), then $a\sqrt3 + b = r(b\sqrt3 + c)$, so $(a-rb)\sqrt3 = rc-b$. But $\sqrt3$ is irrational, and it follows that $a-rb = rc-b = 0$. Therefore $b = rc$ and $a = rb = r^2c$. Then $$\frac{a^2 + b^2 + c^2}{a+b+c} = \frac{r^4c^2 + r^2c^2 + c^2}{r^2c + rc + c} = \frac{(r^4+r^2+1)c}{r^2+r+1}.$$ But $r^4+r^2+1 = (r^2+r+1)(r^2-r+1)$. So $\dfrac{a^2 + b^2 + c^2}{a+b+c} = (r^2-r+1)c = a-b+c$, which is an integer.
 
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