amacorrit80

Answered

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

$\sqrt{2}(pb-qa)=qb-pc$

squaring:

$2(pb-qa{)}^{2}=(qb-pc{)}^{2}$

but that doesn't seems to do much.

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

$\sqrt{2}(pb-qa)=qb-pc$

squaring:

$2(pb-qa{)}^{2}=(qb-pc{)}^{2}$

but that doesn't seems to do much.

Answer & Explanation

Kali Galloway

Expert

2022-07-17Added 16 answers

Let $\frac{a\sqrt{2}+b}{b\sqrt{2}+c}=r\in \mathbb{Q}$

Hence, $(a-rb)\sqrt{2}=rc-b$

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,

$\frac{ab+ac+bc}{a+b+c}=\frac{{b}^{2}(1+r+\frac{1}{r})}{b(1+r+\frac{1}{r})}=b$

and we are done!

Hence, $(a-rb)\sqrt{2}=rc-b$

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,

$\frac{ab+ac+bc}{a+b+c}=\frac{{b}^{2}(1+r+\frac{1}{r})}{b(1+r+\frac{1}{r})}=b$

and we are done!

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