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Math Help - 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, 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!
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  2. #2
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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 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.
    Last edited by Sylvia104; May 4th 2012 at 02:02 PM.
    Thanks from BobtheBob
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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 (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!
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    Re: Prove that a+b+c is a divisor of a^2+b^2+c^2

    \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.
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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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