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Math Help - Prove perfect squares

  1. #1
    Senior Member Mukilab's Avatar
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    Prove perfect squares

    \sqrt{p} +\sqrt{q} +\sqrt{z} =s

    Prove that p, q and r must be perfect squares.

    Ok so the rule for a square is: n^2=(n-1)^2+\sqrt{(2n-1)}

    So p=p-1+\sqrt{2p-1} then p-p=-1+\sqrt{(2p-1)} therefore if you square it 0=1+2p-1 which goes to 0=2p

    Where did I go wrong?
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  2. #2
    Pim
    Pim is offline
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    I think you went wrong in your first step.

    If you substitute n = \sqrt{p} into n^2 = (n-1)^2 + \sqrt{2n-1} you get:

    p = (\sqrt{p}-1)^2 + \sqrt{2 \sqrt{p} - 1} and not what you wrote

    Another thing is that

    (-1 + \sqrt{2p-1})^2 \neq 1 + 2p-1
    it doest equal  (-1 + \sqrt{2p-1})(-1 + \sqrt{2p-1}), where you multiply out the brackets.
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  3. #3
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Mukilab View Post
    Ok so the rule for a square is: n^2=(n-1)^2+\sqrt{(2n-1)}
    Where did you get this? If you plug in n=2, the right hand side yields 1+\sqrt{3} instead of 4.

    From context it's clear that you mean p, q, r, s \in \mathbb{Z}.

    Then it's a matter of proving that if p, q and r are not all perfect squares, then s cannot be rational.
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  4. #4
    Senior Member Mukilab's Avatar
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    That's the formula for proving the square of a number using the previous number squared.
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  5. #5
    Senior Member Mukilab's Avatar
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    I don't know if it's a matter of whether p, q and r are perfect squares or not.

    Never mind, this question does not apply to me anymore and I have far more tricky questions to get around :P
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Mukilab View Post
    That's the formula for proving the square of a number using the previous number squared.
    Well just for reference,

    (n-1)^2=n^2-2n+1

    n^2=(n-1)^2+2n-1
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