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Math Help - 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 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; March 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 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  x=403 .
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    when x = 403, 100x^2-24x-379=16230849
    \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, 100x^2-24x-379=16230849
    \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 100 x^2 - 24 x -379 be a complete square for some integer x > 0?

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

    . . then the equation: 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 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.
     100 x^2 - 24 x -379 = n^2\implies x=\frac{6\pm\sqrt{25n^2+9511}}{50}

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

    So it's equivalent to show the diophantine equation  (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
     100 x^2 - 24 x -379 = n^2\implies x=\frac{6\pm\sqrt{25n^2+9511}}{50}

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

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

    This implies  x=-95 ... but you need  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:  100x^2-24x-379=n^2

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

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

    Factor:  (y+20n)(y-20n)=152176

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

    So you can see, we only have a finite number of cases to consider (  20 to be exact) when solving for  y and  n
    e.g.  \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  y=19024 , or in other words  x=-95 , is the only time there is a solution in  \mathbb{Z} .

    Thus for  x>0 ,  100x^2-24x-379 is never square.
    Last edited by chiph588@; March 28th 2010 at 09:04 PM.
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    Thanks chiph588@
    I was asked to show that 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 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  100x^2-24x-379=n^2 ?
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  11. #11
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    solve y to get y=\frac{1}{2} \left(1+\sqrt{100x^2-24x-379}\right)

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

    Thus 25n^2+9511=m^2 for some positive integer m. That is
    (m+5n)(m-5n)=9511
    Since 9511 is a prime, m+5n = 9511, \quad m-5n = 1
    Thus n=951 but then 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 y=\frac{1}{2} \left(1+\sqrt{100x^2-24x-379}\right)

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

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