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Thread: Diophantine equation question

  1. #1
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    Diophantine equation question

    I've read through a few exercises on Number Theory with regards to Diophantine Equations. Apparently, some of them are solvable and some are not. Here is the example that conflicts me:

    Are there solutions to the Diophantine equation $\displaystyle x^2 + y^2 = 7*z^2$ ? If so, what are they?
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    Quote Originally Posted by Brimley View Post
    I've read through a few exercises on Number Theory with regards to Diophantine Equations. Apparently, some of them are solvable and some are not. Here is the example that conflicts me:

    Are there solutions to the Diophantine equation $\displaystyle x^2 + y^2 = 7*z^2$ ? If so, what are they?
    I don't have time to work on this, but you may find this thread helpful

    http://www.mathhelpforum.com/math-he...ns-148115.html

    And the proof technique of this post (#22) might apply

    http://www.mathhelpforum.com/math-he...tml#post525708
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    undefined, I read that thread you had mentioned and the Wiki articles as well, however those threads had a lot of differing techniques and it was for a specific sample that involved the GCD of the variables equalling 1. I guess what I"m getting at was it looked like you, roninpro, and chip had many different approaches for the same problem, and I tried applying the proof technique of post #22 but didn't get anywhere.

    If you or someone else does have a little time to spare, I would really like to see if there are any sols to this and if so find out what they are.

    Much thanks for any help on this! - Brim
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  4. #4
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    Quote Originally Posted by Brimley View Post
    undefined, I read that thread you had mentioned and the Wiki articles as well, however those threads had a lot of differing techniques and it was for a specific sample that involved the GCD of the variables equalling 1. I guess what I"m getting at was it looked like you, roninpro, and chip had many different approaches for the same problem, and I tried applying the proof technique of post #22 but didn't get anywhere.

    If you or someone else does have a little time to spare, I would really like to see if there are any sols to this and if so find out what they are.

    Much thanks for any help on this! - Brim
    I will try to find time but can't make promises. Off hand though I can tell you that trivially yes there is a solution, (x,y,z) = (0,0,0).
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  5. #5
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    It's simpler to use Fermat's infinite descent:

    Show that the only solution to $\displaystyle x^2+y^2\equiv\: 0\: (\text{mod} \ 7)$ is $\displaystyle x,y\equiv\: 0\: (\text{mod} \ 7)$

    Then suppose $\displaystyle x,y,z$ are solutions to the original problem. From above we get:

    $\displaystyle 49\left(\frac{x}{7}\right)^2+49\left(\frac{y}{7}\r ight)^2=7z^2$

    Therefore $\displaystyle 7\mid z$ and we can write:

    $\displaystyle 49\left(\frac{x}{7}\right)^2+49\left(\frac{y}{7}\r ight)^2=7^3\left(\frac{z}{7}\right)^2$

    $\displaystyle \left(\frac{x}{7}\right)^2+\left(\frac{y}{7}\right )^2=7\left(\frac{z}{7}\right)^2$

    Therefore if $\displaystyle (x,y,z)$ is a solution then so is $\displaystyle \left(\frac{x}{7},\frac{y}{7},\frac{z}{7}\right)$, and applying again and again:

    $\displaystyle \left(\frac{x}{7^n},\frac{y}{7^n},\frac{z}{7^n}\ri ght)$

    is a soloution for all $\displaystyle n\in\mathbb{N}$, and in particular made of integers.

    But it's impossible that these numbers are integers for all $\displaystyle n$ if one of $\displaystyle x,y,z$ is nonzero, so we have the only solution: $\displaystyle (0,0,0)$
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    Quote Originally Posted by Unbeatable0 View Post
    Therefore if $\displaystyle (x,y,z)$ is a solution then so is $\displaystyle \left(\frac{x}{7},\frac{y}{7},\frac{z}{7}\right)$, and applying again and again:

    $\displaystyle \left(\frac{x}{7^n},\frac{y}{7^n},\frac{z}{7^n}\ri ght)$

    is a soloution for all $\displaystyle n\in\mathbb{N}$, and in particular made of integers.

    But it's impossible that these numbers are integers for all $\displaystyle n$ if one of $\displaystyle x,y,z$ is nonzero, so we have the only solution: $\displaystyle (0,0,0)$
    1. What does the $\displaystyle \mathbb{N}$ mean in $\displaystyle n\in\mathbb{N}$ ?
    2. So the only solution is actually the trivial solution of $\displaystyle (0,0,0)$[/QUOTE] ?
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  7. #7
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    Quote Originally Posted by Brimley View Post
    1. What does the $\displaystyle \mathbb{N}$ mean in $\displaystyle n\in\mathbb{N}$ ?
    2. So the only solution is actually the trivial solution of $\displaystyle (0,0,0)$
    1. It means the set of natural numbers (integer numbers greater than or equal to 1)
    2. Indeed
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  8. #8
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    We can generalize this problem:

    Consider the diophantine equation $\displaystyle ax^2+by^2=cz^2$ with $\displaystyle a,b\in\mathbb{Z}\:,\:c\in\mathbb{N}$ satisfying the conditions:

    1) $\displaystyle q$, the "squarefree part" of $\displaystyle c$ (that is, $\displaystyle c=m^2q$ where $\displaystyle q$ is squarefree) is odd.

    2) $\displaystyle a,b$ are coprime to $\displaystyle q$.

    3) $\displaystyle \left(\frac{ab}{q}\right) = (-1)^\frac{q+1}{2}$ (Jacobi Symbol)

    then the only solution to the diophantine equation is the trivial solution.

    Note that in the special case of $\displaystyle ab=\pm 1$, (2) is obviously true and we can replace condition (3) with:

    $\displaystyle (3)_{\pm1}\:\:q \equiv2+ab\pmod{4}$

    (the question which started the thread is an obvious special case of the above)


    Proof:

    $\displaystyle ax^2+by^2=cz^2 \Leftrightarrow ax^2+by^2=q(mz)^2$

    so it's enough to prove that $\displaystyle ax^2+by^2=qz^2$ has no nontrivial solutions.

    Let $\displaystyle (x,y,z)$ be a solution. Then

    $\displaystyle ax^2+by^2 \equiv 0 \pmod{q} \Rightarrow x^2\equiv -a^{-1}by^2 \pmod{q}$.

    Suppose that $\displaystyle \gcd(y,q)\ne 1$, and let $\displaystyle r\mid \gcd(y,q)$ be a prime number. Using the same idea of infinite descent as in my first post in this thread, we get a contradiction.

    Thus $\displaystyle \left(\frac{y}{q}\right)\ne 0 \Rightarrow \left(\frac{y^2}{q}\right) = 1$ and therefore:

    $\displaystyle \left(\frac{-a^{-1}by^2}{q}\right) = \left(\frac{-a^{-1}by^2}{q}\right)\left(\frac{a^2}{q}\right) = \left(\frac{-ab}{q}\right)\left(\frac{y^2}{q}\right) = \left(\frac{-1}{q}\right)\left(\frac{ab}{q}\right)$

    $\displaystyle =(-1)^\frac{q-1}{2}(-1)^\frac{q+1}{2} = -1$

    which contradicts $\displaystyle x^2\equiv -a^{-1}by^2 \pmod{q}$. Therefore $\displaystyle q\mid x$. By symmetry, $\displaystyle q\mid y$, and we can write:

    $\displaystyle aq^2\left(\frac{x}{q}\right)^2+bq^2\left(\frac{y}{ q}\right)^2 = qz^2$.

    (here we refer to the usual fractions, not Jacobi symbols)

    Dividing by $\displaystyle q$ and using the fact that it's squarefree we get $\displaystyle q\mid z$. Then

    $\displaystyle \left(\frac{x}{q},\frac{y}{q},\frac{z}{q}\right)$

    is another solution to the diophantine equation. Continuing ad infinitum yields a contradiction if not all $\displaystyle x,y,z$ are zero.

    We thus get the desired result.
    Last edited by Unbeatable0; Oct 20th 2010 at 09:54 AM. Reason: clarification
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  9. #9
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    Thank you very much unbeatable, i thoroughly enjoyed reading your proof. Thanks again -- Brim
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