If $\frac{a\sqrt{2}+b}{b\sqrt{2}+c}$ is rational, then $a+b+c|ab+bc+ca$

Question

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.

Answer 1

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!