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

• May 4th 2012, 12:08 PM
BobtheBob
Prove that a+b+c is a divisor of a^2+b^2+c^2
Hi,

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

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

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

Thanks very much in advance for any help!
• May 4th 2012, 12:57 PM
Sylvia104
Re: Prove that a+b+c is a divisor of a^2+b^2+c^2
First, note that if one of $\displaystyle a,b$ is $\displaystyle 0,$ then the other also must be $\displaystyle 0.$ If $\displaystyle a=b=0,$ then $\displaystyle a+b+c=c$ is a divisor of $\displaystyle a^2+b^2+c^2=c^2.$

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

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

Finally, we have $\displaystyle (a+b+c)(a-b+c)$ $\displaystyle =$ $\displaystyle (a+c)^2-b^2$ $\displaystyle =$ $\displaystyle a^2+c^2+2ac-b^2$ $\displaystyle =$ $\displaystyle a^2+b^2+c^2$ since $\displaystyle b^2=ca.$
• May 5th 2012, 05:33 AM
BobtheBob
Re: Prove that a+b+c is a divisor of a^2+b^2+c^2
Quote:

Originally Posted by Sylvia104
Rearrange the equation to get $\displaystyle (qa-pb)\sqrt2=pc-qb$. As the RHS is rational, the LHS would be irrational unless $\displaystyle qa-pb=0$. Hence $\displaystyle \frac pq=\frac ab=\frac bc \implies b^2=ca.$

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

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!
• May 5th 2012, 11:29 AM
Sylvia104
Re: Prove that a+b+c is a divisor of a^2+b^2+c^2
$\displaystyle \begin{array} {rcl} \dfrac{a\sqrt2+b}{b\sqrt2+c} &=& \dfrac pq \\\\ qa\sqrt2+qb &=& pb\sqrt2+pc \\\\ (qa-pb)\sqrt2 &=& pc-qb \end{array}$

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

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

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

Actually, by writing $\displaystyle \frac bc,$ I implicitly assumed that $\displaystyle c\ne0.$ To make absolutely sure, I should check that if $\displaystyle c=0,$ then $\displaystyle \frac{a\sqrt2+b}{b\sqrt2+c}=\frac ab+\frac1\sqrt2$ which is not rational; therefore $\displaystyle \frac{a\sqrt2+b}{b\sqrt2+c}$ rational $\displaystyle \implies$ $\displaystyle c\ne0.$
• May 6th 2012, 07:01 AM
BobtheBob
Re: Prove that a+b+c is a divisor of a^2+b^2+c^2
Oh right! Thanks so much for the help! It always seems so simple when you know how.

Thanks again.