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I How to find if a quadratic expression is a perfect square

  1. Jul 27, 2016 #1
    How to find if a quadratic expression of the form
    4x2 + 4.n.x - P ..........................(x,n and P are natural number)
    is a perfect square.

    For example,
    4x2 + 64x - 31

    Thanks.
     
  2. jcsd
  3. Jul 27, 2016 #2

    mfb

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    If it is a perfect square it can be written as a(x+d)2 = a x2 + 2 a d x + ad2. You can compare the individual summands to determine "a" and "d" and see if all three components fit. For example, take the x2 term to determine a.

    Edit: Moved to variables a and d to avoid confusion with other uses of b.
     
    Last edited: Jul 27, 2016
  4. Jul 27, 2016 #3

    micromass

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    Or you can compute the discriminant ##D = b^2 - 4ac##.
     
  5. Jul 27, 2016 #4
    Thanks. I mean
    4x2 + 64x - 31

    will generate
    37
    113
    197
    289 (perfect square, for x=4)
    .. and so on.

    I want to know is there anyway if a quadratic expression will ever generate a perfect square for some 'x' value.
     
  6. Jul 27, 2016 #5

    micromass

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    Then you'll need to find ##n## and ##m## such that
    [tex]4n^2 + 64n - 31 = m^2[/tex]

    I have solved very similar questions already in your previous threads. So I'm sure you can handle it from here.
     
  7. Sep 10, 2016 #6
    I have made a slight change, say now the quadratic expression is ( I have only removed the coefficient of x^2)

    x^2 + 648x + 247 = m^2, ( x, m both are integer). Also for some purpose I will be considering quadratic expression where the 'x' term and the constant(here 247) are either odd-even or viceversa.

    I can break the above quadratic expression into,

    x^2 + ( 2*323*x + 2x ) + ( 323^2 - 2y ) ....(1)
    x^2 + ( 2*322*x + 4x ) + ( 322^2 - 4y ) ....(2)
    x^2 + ( 2*321*x + 6x ) + ( 321^2 - 6y ) ....(3)

    and so on ... untill
    x^2 + ( 2*1*x + 2*323*x ) + ( 1^2 - 2*323*y ) ...(323)

    Lets take equ(1)
    it can be rearrange into
    ( x^2 + 2*323*x + 323^2 ) + 2(x-y)
    (x + 323 )^2 + 2(x-y)

    So if x=y, then the above quadratic expression has a perfect square. But since we are only taking x and m as integer, here too y has to be an integer.

    Also,
    323^2 - 2y = 247
    2y = ( 323^2 - 247 )
    y = ( 323^2 - 247 ) / 2

    which is divisible by 2
    i.e y = 52041 ( an integer)

    For every such quadratic expression (in the general form, where the x term and the constant are either odd-even or viceversa)
    dividing with 2(i.e first equation equ(1) ) is always a perfect square. Which I do not need in my problem.

    Similarly for equ(2,3,...), it is
    y = ( 322^2 - 247 ) / 4
    y = ( 321^2 - 247 ) / 6
    y = ( 320^2 - 247 ) / 8 and so on...

    If any on the above y value is an integer then the above quadratic expression will have as many perfect square.

    But I cannot do them all these manually, I failed to get other any way that can tell me if the quadratic can generate perfect square(ignoring dividing by 2).

    Thank you.
     
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