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Thread: For any integer x, 100 x^2 - 24 x - 379 is not a complete square

  1. #1
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    For any integer x, 100 x^2 - 24 x - 379 is not a complete square

    Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

    Sorry I should add a condition x > 0 in the subject line.
    Last edited by elim; Mar 27th 2010 at 07:57 AM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

    Sorry I should add a condition x > 0 in the subject line.
    Try $\displaystyle x=403 $.
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    when x = 403, $\displaystyle 100x^2-24x-379=16230849$
    $\displaystyle \sqrt{16230849}=4028.752784671\cdots$
    Not work.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    when x = 403, $\displaystyle 100x^2-24x-379=16230849$
    $\displaystyle \sqrt{16230849}=4028.752784671\cdots$
    Not work.
    Whoops! I typed something in wrong earlier. My mistake!
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  5. #5
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    Hello, elim!

    Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

    If $\displaystyle 100x^2 - 24x - 379$ is the square of any real number,

    . . then the equation: $\displaystyle x^2-24x - 379 \:=\:0$ has a discriminant equal to 0.


    Does it?

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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

    Sorry I should add a condition x > 0 in the subject line.
    $\displaystyle 100 x^2 - 24 x -379 = n^2\implies x=\frac{6\pm\sqrt{25n^2+9511}}{50} $

    Thus we require $\displaystyle 25n^2+9511 $ to be square to give $\displaystyle x $ a fighting chance to be an integer.

    So it's equivalent to show the diophantine equation $\displaystyle (5n)^2-m^2=-9511 $ has no solution.
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    $\displaystyle 100 x^2 - 24 x -379 = n^2\implies x=\frac{6\pm\sqrt{25n^2+9511}}{50} $

    Thus we require $\displaystyle 25n^2+9511 $ to be square to give $\displaystyle x $ a fighting chance to be an integer.

    So it's equivalent to show the diophantine equation $\displaystyle (5n)^2-m^2=-9511 $ has no solution.
    A solution to $\displaystyle (5n)^2-m^2=-9511 $ is $\displaystyle (n,m)=(951,4756) $.

    This implies $\displaystyle x=-95 $... but you need $\displaystyle x>0 $.

    This is a really tough problem! What is this for?
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    We want to solve: $\displaystyle 100x^2-24x-379=n^2 $

    Complete the square by multiplying both sides by $\displaystyle 400 $ and adding and subtracting $\displaystyle 24^2 $: $\displaystyle (200x-24)^2-400n^2-152176=0 $

    Let $\displaystyle y=200x-24 $: $\displaystyle y^2-400n^2-152176=0 $

    Factor: $\displaystyle (y+20n)(y-20n)=152176 $

    Now let's factor $\displaystyle 152176 $: $\displaystyle 152176 = 2^4\cdot9511 $

    So you can see, we only have a finite number of cases to consider ($\displaystyle 20 $ to be exact) when solving for $\displaystyle y $ and $\displaystyle n $
    e.g. $\displaystyle \begin{cases} y+20n=4 \\ y-20n=38044 \end{cases} $ is one example.

    I'll leave you to consider all these cases, but you should find that $\displaystyle y=19024 $, or in other words $\displaystyle x=-95 $, is the only time there is a solution in $\displaystyle \mathbb{Z} $.

    Thus for $\displaystyle x>0 $, $\displaystyle 100x^2-24x-379 $ is never square.
    Last edited by chiph588@; Mar 28th 2010 at 09:04 PM.
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  9. #9
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    Thanks chiph588@
    I was asked to show that $\displaystyle 25x^2-6x-95=y(y-1)$ has no positive integer solutions.

    And this leads to the problem here.
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    Thanks chiph588@
    I was asked to show that $\displaystyle 25x^2-6x-95=y(y-1)$ has no positive integer solutions.

    And this leads to the problem here.
    How did you reduce this problem to $\displaystyle 100x^2-24x-379=n^2 $?
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  11. #11
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    solve y to get $\displaystyle y=\frac{1}{2} \left(1+\sqrt{100x^2-24x-379}\right)$

    Then as you pointed, $\displaystyle x = (6+\sqrt{25n^2+9511})/50$

    Thus $\displaystyle 25n^2+9511=m^2$ for some positive integer $\displaystyle m$. That is
    $\displaystyle (m+5n)(m-5n)=9511$
    Since 9511 is a prime, $\displaystyle m+5n = 9511, \quad m-5n = 1$
    Thus $\displaystyle n=951$ but then $\displaystyle x$ is not an integer.
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  12. #12
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    solve y to get $\displaystyle y=\frac{1}{2} \left(1+\sqrt{100x^2-24x-379}\right)$

    Then as you pointed, $\displaystyle x = (6+\sqrt{25n^2+9511})/50$

    Thus $\displaystyle 25n^2+9511=m^2$ for some positive integer $\displaystyle m$. That is
    $\displaystyle (m+5n)(m-5n)=9511$
    Since 9511 is a prime, $\displaystyle m+5n = 9511, \quad m-5n = 1$
    Thus $\displaystyle n=951$ but then $\displaystyle x$ is not an integer.
    This is better than my solution since there are much less cases to consider!
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