amacorrit80

2022-07-16

If $\frac{a\sqrt{2}+b}{b\sqrt{2}+c}$ is rational, then $a+b+c|ab+bc+ca$
Let $a,b,c\in {\mathbb{N}}_{\mathbb{>}\mathbb{0}}$. 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
$\sqrt{2}\left(pb-qa\right)=qb-pc$
squaring:
$2\left(pb-qa{\right)}^{2}=\left(qb-pc{\right)}^{2}$
but that doesn't seems to do much.

Kali Galloway

Expert

Let $\frac{a\sqrt{2}+b}{b\sqrt{2}+c}=r\in \mathbb{Q}$
Hence, $\left(a-rb\right)\sqrt{2}=rc-b$
If $a-rb\ne 0$ 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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