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Thread: Prove perfect squares

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

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

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

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

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

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

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

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

    Another thing is that

    $\displaystyle (-1 + \sqrt{2p-1})^2 \neq 1 + 2p-1$
    it doest equal $\displaystyle (-1 + \sqrt{2p-1})(-1 + \sqrt{2p-1})$, 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: $\displaystyle n^2=(n-1)^2+\sqrt{(2n-1)}$
    Where did you get this? If you plug in $\displaystyle n=2$, the right hand side yields $\displaystyle 1+\sqrt{3}$ instead of $\displaystyle 4$.

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

    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,

    $\displaystyle (n-1)^2=n^2-2n+1$

    $\displaystyle n^2=(n-1)^2+2n-1$
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