Results 1 to 4 of 4

Thread: Extra Solutions of a Quadratic Diophantine Equation

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1

    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.)
    $\displaystyle a^2 + a + 2 = 2c^2$

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

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

    $\displaystyle (2a + 1)^2 + 7 = 8c^2$

    Let b = 2a + 1

    $\displaystyle b^2 + 7 = 8c^2$

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

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

    so
    $\displaystyle b = 7y \pm c$

    where y is any integer.

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

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

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

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

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

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

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

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Actually $\displaystyle (x,y)=(3,1)$ generates $\displaystyle (a,c)=(2,2)$

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

    You might be interested in reading here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by PaulRS View Post
    Actually $\displaystyle (x,y)=(3,1)$ generates $\displaystyle (a,c)=(2,2)$

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

    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:

    $\displaystyle b^2 - 8c^2 = -7$

    $\displaystyle b = 7y \pm c$

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Actually, $\displaystyle (a,c)=(2,2)$ is comes from:

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

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

    I prefer going though linear equations though:

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

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

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

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

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

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

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

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

    Here with the linear equation we don't need to consider $\displaystyle (\pm b,\pm c)$ since the different solutions come from $\displaystyle (\pm x,\pm y)$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic Diophantine equation
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Jan 23rd 2011, 11:43 AM
  2. Positive Solutions to Linear Diophantine Equation
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Aug 3rd 2010, 11:22 PM
  3. How does one solve a quadratic Diophantine equation?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 17th 2009, 04:46 AM
  4. Solutions to this diophantine equation.
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Oct 25th 2008, 01:19 PM
  5. Finding all solutions to a Diophantine equation
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Oct 10th 2006, 01:10 PM

Search Tags


/mathhelpforum @mathhelpforum