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Problem Of The Week #439 October 19th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
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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