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

  1. #1
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    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!
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  2. #2
    Member Sylvia104's Avatar
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    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 $\displaystyle b^2=ca.$
    Last edited by Sylvia104; May 4th 2012 at 01:02 PM.
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  3. #3
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    Re: Prove that a+b+c is a divisor of a^2+b^2+c^2

    Quote Originally Posted by Sylvia104 View Post
    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!
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  4. #4
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    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 $\displaystyle \implies$ $\displaystyle c\ne0.$
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  5. #5
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    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.
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