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Math Help - Extra Solutions of a Quadratic Diophantine Equation

  1. #1
    Forum Admin topsquark's Avatar
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    Extra Solutions of a Quadratic Diophantine Equation

    Okay, I've been making a brief foray into quadratic Diophantine equations. Here's my practice equation and it's solution to Pell form. (My questions will come after the derivation.)
    a^2 + a + 2 = 2c^2

    a^2 + a + \frac{1}{4} - \frac{1}{4} + 2 = 2c^2

    \left ( a + \frac{1}{2} \right ) ^2 + \frac{7}{4} = 2c^2

    (2a + 1)^2 + 7 = 8c^2

    Let b = 2a + 1

    b^2 + 7 = 8c^2

    Now take this a modulo 7:
    b^2 \equiv c^2~\text{(mod 7)}

    Thus, since 7 is prime we obtain:
    b \equiv \pm c

    so
    b = 7y \pm c

    where y is any integer.

    Plugging this in for b:
    (7y \pm c)^2 + 7 = 8c^2

    49y^2 \pm 14yc + c^2 + 7 = 8c^2

    49y^2 \pm 14yc - 7c^2 = -7

    7y^2 \pm 2yc - c^2 = -1

    c^2 \mp 2yc + y^2 - y^2 - 7y^2 = 1

    (c \mp y)^2 - 8y^2 = 1

    Let x = c \mp y giving (finally!)
    x^2 - 8y^2 = 1

    This is a Pell equation. One solution to this is (x, y) = (1, 0), which generates the solutions (a, c) = (0, 1); (-1, 1).

    A second solution (x, y) = (-1, 0) generates (a, c) = (0, -1); (-1, -1). So far so good.

    Another solution to the Pell equation is (x, y) = (3, 1), which generates the solutions (a, c) = (4, 2); (5, 4). We can generate (a, c) = (4, -2); (5, -4) from (x, y) = (-3, 1). ((x, y) = (3, -1) and (-3, -1) simply copy these solutions.)

    My first question: The "solutions" (a, c) = (4, 2); (4, -2) do not solve the original equation. Why do these solutions appear?

    My second question: Two solutions (a, c) = (2, -2); (2, 2) do not appear using this method. Yet I understood that all solutions to the original equation derive from solutions to the Pell equation? I can't find a solution to the Pell equation that generates them.

    Thanks!

    -Dan
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  2. #2
    Super Member PaulRS's Avatar
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    Actually (x,y)=(3,1) generates (a,c)=(2,2)

    Indeed c = x - y = 2 and b = 7\cdot y - c = 5 (and the signs do make sense) thus a = \frac{b-1}{2} = 2

    You might be interested in reading here.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by PaulRS View Post
    Actually (x,y)=(3,1) generates (a,c)=(2,2)

    Indeed c = x - y = 2 and b = 7\cdot y - c = 5 (and the signs do make sense) thus a = \frac{b-1}{2} = 2

    You might be interested in reading here.
    I will look at the document. Thank you!

    I'm still having a problem with the derivation of (a, c) = (2, 2) from (x, y) = (3, 1). Keeping track of the signs was confusing at first, so I did the problem backward. We need the equations:

    b^2 - 8c^2 = -7

    b = 7y \pm c

    Now, the solution for c from the first equation is what gives the (+/-) in the second equation. When I plug that through the first solution that only gives me two possibilities for (x, y) and neither solve the Pell equation. In order to generate a valid solution to the Pell equation, that is (a, c) = (2, 2), we need to assume that the (+/-) in the second equation is an extra (+/-) not correlated to the (+/-) on the solution of c. I can't see how this can be.

    -Dan
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  4. #4
    Super Member PaulRS's Avatar
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    Actually, (a,c)=(2,2) is comes from:

    b^2-8c^2 = -7 and b=7\cdot 1 - c ...

    Don't forget that if (b,c) is a solution to b^2-8c^2=-7, then so is (\pm b,\pm c) (with all 4 possibilities)

    I prefer going though linear equations though:

    You have (look back at your equations) : x = c \mp y and b = 7\cdot y \pm c (you use the signs that are that the same level )

    Thus:
    c = x \pm y and b = 7\cdot y \pm c : only 2 possibilities (either plus in both, or minus in both).

    And then a = \frac{b-1}{2}.

    With (x,y)=(3,1) we get:

    c = 3 \pm 1 and b = 7\cdot 1 \pm c

    Pick the - sign then : c = 2 and b = 5 thus a = 2

    Pick the + sign then: c = 4 and b = 7\cdot 1 + 4 = 11 hence a = 5

    Both (a,c)=(2,2) and (a,c)=(5,4) are solutions to the original equation.

    Here with the linear equation we don't need to consider (\pm b,\pm c) since the different solutions come from (\pm x,\pm y).
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