Prove that a+b+c is a divisor of a^2+b^2+c^2

**BobtheBob** · May 4th 2012, 12:08 PM

Hi,

I was hoping somebody could put me on the right line to prove this question. It says:

Let, $a, b, c$ be positive integers such that $\frac{a \sqrt{2} + b}{b \sqrt{2} + c}$ is a rational number. Prove that $a+b+c$ is a divisor of $a^2+b^2+c^2$

I'm never very good at proofs as I never know where to start. Would one times $\frac{a \sqrt{2} + b}{b \sqrt{2} + c}$ by $a+b+c$ and see what happens (I have but I can't see anything helpful)?

Thanks very much in advance for any help!

**Sylvia104** · May 4th 2012, 12:57 PM

First, note that if one of $a,b$ is $0,$ then the other also must be $0.$ If $a=b=0,$ then $a+b+c=c$ is a divisor of $a^2+b^2+c^2=c^2.$

Assume $ab\ne0$ and let $\frac{a\sqrt2+b}{b\sqrt2+c}=\frac pq$ where $p,q\in\mathbb Z$ and $q\ne0.$

Rearrange the equation to get $(qa-pb)\sqrt2=pc-qb.$ As the RHS is rational, the LHS would be irrational unless $qa-pb=0.$ Hence $\frac pq=\frac ab=\frac bc$ $\implies$ $b^2=ca.$

Finally, we have $(a+b+c)(a-b+c)$ $=$ $(a+c)^2-b^2$ $=$ $a^2+c^2+2ac-b^2$ $=$ $a^2+b^2+c^2$ since $b^2=ca.$

**BobtheBob** · May 5th 2012, 05:33 AM

Originally Posted by Sylvia104

Rearrange the equation to get $(qa-pb)\sqrt2=pc-qb$ . As the RHS is rational, the LHS would be irrational unless $qa-pb=0$ . Hence $\frac pq=\frac ab=\frac bc \implies b^2=ca.$

Finally, we have $(a+b+c)(a-b+c) = (a+c)^2-b^2 = a^2+c^2+2ac-b^2 = a^2+b^2+c^2$ since $b^2=ca$

Ok right I sort-of see what's happening. Although I'm not totally sure on some steps above.

Could you just expalin the method a little more as I am slightly unsure of what you are doing?

Thank you very much for your help so far!

**Sylvia104** · May 5th 2012, 11:29 AM

$\begin{array} {rcl} \dfrac{a\sqrt2+b}{b\sqrt2+c} &=& \dfrac pq \\\\ qa\sqrt2+qb &=& pb\sqrt2+pc \\\\ (qa-pb)\sqrt2 &=& pc-qb \end{array}$

$\therefore\ qa-pb=pc-qb=0$

$qa=pb\ \text{and}\ pc=qb$

$\frac pq=\frac ab\ \text{and}\ \frac pq =\frac bc$

Actually, by writing $\frac bc,$ I implicitly assumed that $c\ne0.$ To make absolutely sure, I should check that if $c=0,$ then $\frac{a\sqrt2+b}{b\sqrt2+c}=\frac ab+\frac1\sqrt2$ which is not rational; therefore $\frac{a\sqrt2+b}{b\sqrt2+c}$ rational $\implies$ $c\ne0.$

**BobtheBob** · May 6th 2012, 07:01 AM

Oh right! Thanks so much for the help! It always seems so simple when you know how.

Thanks again.

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