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Math Help - Diophantine equation

  1. #1
    MHF Contributor alexmahone's Avatar
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    Diophantine equation

    Find all integral solutions of the equation x^4+2x^3+2x^2+2x+5=y^2.

    My attempt: The equation can be rewritten as (x+1)^2(x^2+1)=(y+2)(y-2).

    Can someone please tell me how to proceed?
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  2. #2
    Member Pranas's Avatar
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    Quote Originally Posted by alexmahone View Post
    Find all integral solutions of the equation x^4+2x^3+2x^2+2x+5=y^2.

    My attempt: The equation can be rewritten as (x+1)^2(x^2+1)=(y+2)(y-2).

    Can someone please tell me how to proceed?
    I hate to be writing the very first comment in this thread without any actual help within, but it has been resting for a while now...

    I noticed that

    \displaystyle \[{x^4} + 2{x^3} + 2{x^2} + 2x + 5 = {y^2}\]

    May be rewritten as (quite unusual form)

    \displaystyle \[x\left( {x\left( {x\left( {x + 2} \right) + 2} \right) + 2} \right) + 5 = {y^2}\]

    I don't have a clue whether there can be a slightest use of this though, sorry...
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  3. #3
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    Re-wrote it as:

     <br />
(x+1)(x+1)(x-1)(x-1)(0.5x)=(y+2)(y-2)<br /> <br />
...

    <br /> <br />
(x^2-1)^2(0.5x)=(y^2-4)<br /> <br />
...

    <br /> <br />
(x^2-1)^2=(y^2-4)/(0.5x)<br /> <br />
...

    <br /> <br />
x((x^2-1)^2)=(y^2-4)/(0.5)<br /> <br />
...

    <br /> <br />
(x)(x^2-1)(x^2-1)=(2)(y+2)(y-2)<br /> <br />
...

    <br /> <br />
(x)(x-1)(x+1)(x-1)(x+1)=(2)(y+2)(y-2)<br /> <br />
...

    <br /> <br />
(x)(x^2+2x+1)(x^2-2x-1)=(2)(y+2)(y-2)<br /> <br />
...

    <br /> <br />
(xxxxx-4xxx-4xx-x)=(2y+4)(2y-4)<br /> <br />
...

    <br /> <br />
xxxxx-4xxx-4xx-x=4yy-16<br />


    I don't know where I'm going with this, but maybe that helped
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  4. #4
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    Quote Originally Posted by orange gold View Post
    Re-wrote it as:

     <br />
(x+1)(x+1)(x-1)(x-1)(0.5x)=(y+2)(y-2)...


    Your left side here doesn't equal the OP's left side: it must be a quartic pol. in x, and you wrote a quintic.

    Tonio





    <br /> <br />
(x^2-1)^2(0.5x)=(y^2-4)<br /> <br />
...

    <br /> <br />
(x^2-1)^2=(y^2-4)/(0.5x)<br /> <br />
...

    <br /> <br />
x((x^2-1)^2)=(y^2-4)/(0.5)<br /> <br />
...

    <br /> <br />
(x)(x^2-1)(x^2-1)=(2)(y+2)(y-2)<br /> <br />
...

    <br /> <br />
(x)(x-1)(x+1)(x-1)(x+1)=(2)(y+2)(y-2)<br /> <br />
...

    <br /> <br />
(x)(x^2+2x+1)(x^2-2x-1)=(2)(y+2)(y-2)<br /> <br />
...

    <br /> <br />
(xxxxx-4xxx-4xx-x)=(2y+4)(2y-4)<br /> <br />
...

    <br /> <br />
xxxxx-4xxx-4xx-x=4yy-16<br />


    I don't know where I'm going with this, but maybe that helped
    .
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  5. #5
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by alexmahone View Post
    Find all integral solutions of the equation x^4+2x^3+2x^2+2x+5=y^2.

    My attempt: The equation can be rewritten as (x+1)^2(x^2+1)=(y+2)(y-2).

    Can someone please tell me how to proceed?
    Let p(x) = x^4+2x^3+2x^2+2x+5. Check that

    (1)\qquad (x^2+x)^2 = p(x) - (x^2+2x+5),

    (2)\qquad (x^2+x+1)^2 = p(x) - (x^2-4),

    (3)\qquad (x^2+x+2)^2 = p(x) + (3x^2+2x-1).

    Then x^2+2x+5 = (x+1)^2+4>0, so it follows from (1) that x^2+x is always too small to be a square root for p(x).

    Next, 3x^2+2x-1 = (3x-1)(x+1), and the only integers for which this is not positive are x=0 and x=-1. For all other integers, 3x^2+2x-1 is positive, and it follows from (3) that x^2+x+2 is too large to be a square root of p(x).

    Thus, unless x = 0 or -1, the only possible candidate for an integral square root of p(x) is y = x^2+x+1, and it follows from (2) that this solution will only work if x^2-4=0.

    Therefore the only values of \normalsize x that might give solutions to the problem are x = 0,\ -1 and \pm2. If \normalsize x=0 then p(x) = 5, which is not a square. But the other three values of \normalsize x give the solutions to the problem, namely (x,y) = (-2,\pm3),\ (-1,\pm2),\ (2,\pm7).
    Last edited by Opalg; January 16th 2011 at 01:58 AM.
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  6. #6
    Senior Member roninpro's Avatar
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    Quote Originally Posted by Opalg View Post
    Let p(x) = x^4+2x^3+2x^2+2x+5. Check that

    (1)\qquad (x^2+x)^2 = p(x) - (x^2+2x+5),

    (2)\qquad (x^2+x+1)^2 = p(x) - (x^2-4),

    (3)\qquad (x^2+x+2)^2 = p(x) + (3x^2+2x-1).

    Then x^2+2x+5 = (x+1)^2+4>0, so it follows from (1) that x^2+x is always too small to be a square root for p(x).

    Next, 3x^2+2x-1 = (3x-1)(x+1), and the only integers for which this is not positive are x=0 and x=-1. For all other integers, 3x^2+2x-1 is positive, and it follows from (3) that x^2+x+2 is too large to be a square root of p(x).

    Thus, unless x = 0 or -1, the only possible candidate for an integral square root of p(x) is y = x^2+x+1, and it follows from (2) that this solution will only work if x^2-4=0.

    Therefore the only values of \normalsize x that might give solutions to the problem are x = 0,\ -1 and \pm2. If \normalsize x=0 then p(x) = 5, which is not a square. But the other three values of \normalsize x give the solutions to the problem, namely (x,y) = (-2,\pm3),\ (-1,\pm2),\ (2,\pm7).
    Could you elaborate a little bit more on why this method is exhaustive?
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  7. #7
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    Quote Originally Posted by roninpro View Post
    Quote Originally Posted by Opalg View Post
    Let p(x) = x^4+2x^3+2x^2+2x+5. Check that

    (1)\qquad (x^2+x)^2 = p(x) - (x^2+2x+5),

    (2)\qquad (x^2+x+1)^2 = p(x) - (x^2-4),

    (3)\qquad (x^2+x+2)^2 = p(x) + (3x^2+2x-1).

    Then x^2+2x+5 = (x+1)^2+4>0, so it follows from (1) that x^2+x is always too small to be a square root for p(x).

    Next, 3x^2+2x-1 = (3x-1)(x+1), and the only integers for which this is not positive are x=0 and x=-1. For all other integers, 3x^2+2x-1 is positive, and it follows from (3) that x^2+x+2 is too large to be a square root of p(x).

    Thus, unless x = 0 or -1, the only possible candidate for an integral square root of p(x) is y = x^2+x+1, and it follows from (2) that this solution will only work if x^2-4=0.

    Therefore the only values of \normalsize x that might give solutions to the problem are x = 0,\ -1 and \pm2. If \normalsize x=0 then p(x) = 5, which is not a square. But the other three values of \normalsize x give the solutions to the problem, namely (x,y) = (-2,\pm3),\ (-1,\pm2),\ (2,\pm7).
    Could you elaborate a little bit more on why this method is exhaustive?
    The strategy is quite simple really. We want to find an integer that is a square root for p(x). My claim is that the best candidate for this square root is z \stackrel{\mathrm{d{e}f}}= x^2+x+1. Equation (1) shows that z-1 is definitely too small to be a square root for p(x). Similarly, equation (3) shows that z+1 is too large to be a square root for p(x) except possibly when x = -1 or 0. So, for all other values of \normalsize x, the square root must be bigger than z-1 and smaller than z+1, which leaves  z as the only remaining possibility. But equation (2) shows that for  z to be the square root, it is necessary that x^2-4=0, and that only happens when x=\pm2.
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Opalg View Post
    The strategy is quite simple really. We want to find an integer that is a square root for p(x). My claim is that the best candidate for this square root is z \stackrel{\mathrm{d{e}f}}= x^2+x+1. Equation (1) shows that z-1 is definitely too small to be a square root for p(x). Similarly, equation (3) shows that z+1 is too large to be a square root for p(x) except possibly when x = -1 or 0. So, for all other values of \normalsize x, the square root must be bigger than z-1 and smaller than z+1, which leaves  z as the only remaining possibility. But equation (2) shows that for  z to be the square root, it is necessary that x^2-4=0, and that only happens when x=\pm2.
    Just to pick your brain a bit, what led you to choose  z=x^2+x+1 in the first place?

    Edit: Wait, I think I see why.  z^2=x^4+2x^3+3x^2+2x+1 \approx p(x)
    Last edited by chiph588@; January 16th 2011 at 03:34 PM.
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