<s>[INST] If $\frac{a\sqrt{2}+b}{b\sqrt{2}+c}$ is rational, then $a+b+c|ab+bc+ca$ Let $a,b,c\in\mathbb{N_{>0}}$. Prove that if $$\frac{a\sqrt{2}+b}{b\sqrt{2}+c}$$ is a rational number, then $a+b+c|ab+bc+ca$.
Hints? 
My approach was to assume that
$$ \frac{a\sqrt{2}+b}{b\sqrt{2}+c}=\frac{p}{q}$$
then
$$ pb\sqrt{2}+pc=qa\sqrt{2}+qb$$
or 
$$ \sqrt{2}(pb-qa)=qb-pc$$
squaring:
$$ 2(pb-qa)^2=(qb-pc)^2$$
but that doesn't seems to do much. 
 [/INST] Let $\frac{a\sqrt2+b}{b\sqrt2+c}=r\in\mathbb Q$.
Hence, $(a-rb)\sqrt2=rc-b$.
If $a-rb\neq0$ so we get a contradiction.
Thus, $a-rb=0$ and $rc-b=0$ or $a=rb$ and $c=\frac{b}{r}$.
Id est, $$\frac{ab+ac+bc}{a+b+c}=\frac{b^2\left(1+r+\frac{1}{r}\right)}{b\left(1+r+\frac{1}{r}\right)}=b$$
and we are done!
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