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  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 x^2 + y^2 = 7*z^2 ? If so, what are they?
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  2. #2
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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 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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  3. #3
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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 x^2+y^2\equiv\: 0\: (\text{mod} \ 7) is x,y\equiv\: 0\: (\text{mod} \ 7)

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

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

    Therefore 7\mid z and we can write:

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

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

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

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

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

    But it's impossible that these numbers are integers for all n if one of x,y,z is nonzero, so we have the only solution: (0,0,0)
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  6. #6
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    Quote Originally Posted by Unbeatable0 View Post
    Therefore if (x,y,z) is a solution then so is \left(\frac{x}{7},\frac{y}{7},\frac{z}{7}\right), and applying again and again:

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

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

    But it's impossible that these numbers are integers for all n if one of x,y,z is nonzero, so we have the only solution: (0,0,0)
    1. What does the \mathbb{N} mean in n\in\mathbb{N} ?
    2. So the only solution is actually the trivial solution of (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 \mathbb{N} mean in n\in\mathbb{N} ?
    2. So the only solution is actually the trivial solution of (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 ax^2+by^2=cz^2 with a,b\in\mathbb{Z}\:,\:c\in\mathbb{N} satisfying the conditions:

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

    2) a,b are coprime to q.

    3) \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 ab=\pm 1, (2) is obviously true and we can replace condition (3) with:

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

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


    Proof:

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

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

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

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

    Suppose that \gcd(y,q)\ne 1, and let 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 \left(\frac{y}{q}\right)\ne 0 \Rightarrow \left(\frac{y^2}{q}\right) = 1 and therefore:

    \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)

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

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

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

    \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 x,y,z are zero.

    We thus get the desired result.
    Last edited by Unbeatable0; October 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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