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Math Help - yet another Diophantine equation

  1. #1
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    yet another Diophantine equation

    Hi again,

    Have struggled with this for a while but getting nowhere.

    x^2=y^2-2yz-z^2

    I've reduced this to

     x^2=(y-z)^2-2z^2

    but can't seem to break it down any further in a similar method to my previous thread

    http://www.mathhelpforum.com/math-he...tml#post674932

    Any ideas would be greatly appreciated
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  2. #2
    Grand Panjandrum
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    Re: yet another Diophantine equation

    Quote Originally Posted by procyon View Post
    Hi again,

    Have struggled with this for a while but getting nowhere.

    x^2=y^2-2yz-z^2

    I've reduced this to

     x^2=(y-z)^2-2z^2

    but can't seem to break it down any further in a similar method to my previous thread

    http://www.mathhelpforum.com/math-he...tml#post674932

    Any ideas would be greatly appreciated
    Put w=y-z then you have:

    w^2-x^2=2z^2

    CB
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  3. #3
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    Re: yet another Diophantine equation

    Quote Originally Posted by CaptainBlack View Post
    Put w=y-z then you have:

    w^2-x^2=2z^2

    CB
    Thanks for the reply Captain. I probably should have added that for clarity but thought since I showed

    this^2=that^2-2(theOther)^2

    would be enough.

    It's a way of finding integer roots of this (not all, but not by trial & error either) I'm trying to get, similar to the last post I mentioned

    http://www.mathhelpforum.com/math-he...tml#post674932

    Pro
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  4. #4
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    Re: yet another Diophantine equation

    Another thing about this, I can find no integer solutions when z is odd.
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  5. #5
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    Re: yet another Diophantine equation

    Quote Originally Posted by procyon View Post
    Another thing about this, I can find no integer solutions when z is odd.
    There are no integer solutions when z is odd, to see this (and how to find the other solutions) write:

    w^2-x^2=(w+x)(w-x)=2z^2

    Then for any factorisation z_1z_2 of z^2 we may set

    w+x=2z_1

    w-x=z_2

    If z_2 is odd then from the second of these equations w and x are of different parities, but from the first they are of the same parity. This is a contradiction so for there to a solution z_2 must be even, which is not possible if z (and also z^2) is odd.

    CB
    Last edited by CaptainBlack; September 7th 2011 at 06:14 AM. Reason: fix typo
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  6. #6
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    Re: yet another Diophantine equation

    Many thanks for the above CB. I had been working at it a more difficult way as usual, lol.

    I'm not sure if I have made myself clear on what it is I actually seek.

    I want to be able to reduce

    w^2-x^2=2z^2

    into some form of

    a^2+b^2=c^2

    even if it is like this

    \left(\dfrac{d+e}{f}\right)^2+\left(\dfrac{g+h}{i}  \right)^2=\left(\dfrac{j+k}{l}\right)^2


    I'll start here x^2+2z^2=w^2

    I have used the fact that z must be even and that w and x must both be even or both be odd to no avail

    i.e.

    z is even and so there is some integer p>0 such that z=2p

    let w and x both be even, so similarly, w=2q and x=2r

    so we now have 4r^2+8p^2=4q^2 \Leftrightarrow q^2-r^2=2p^2

    which is back to the start

    let's forget that z must be even but still have w and z such

    4r^2+2z^2=4q^2\Leftrightarrow 2r^2+z^2=2q^2

    It's different, but still not what I'm after.
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  7. #7
    MHF Contributor alexmahone's Avatar
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    Re: yet another Diophantine equation

    Quote Originally Posted by CaptainBlack View Post
    There are no integer solutions when z is odd, to see this (and how to find the other solutions) write:

    w^2-x^2=(w+x)(w-x)=2z^2

    Then for any factorisation z_1z_2 of z we may set

    w+x=2z_1

    w-x=z_2

    If z_2 is odd then from the second of these equations w and x are of different parities, but from the first they are of the same parity. This is a contradiction so for there to a solution z_2 must be even, which is not possible if z is odd.

    CB
    Surely, you mean:

    w+x=2z_1^2 ---------- (1)

    w-x=z_2^2 ---------- (2)

    ------------------------------------------------------------------------------------------

    Let z_1=1 and z_2=2

    z=z_1z_2=2

    Solving (1) and (2), we get

    w=3 and x=-1

    y-z=3

    y-2=3

    y=5

    So, (-1, 5, 2) is a solution.

    Note that (1, 5, 2) is also a solution. (This is obtained by setting w+x=z_2^2 and w-x=2z_1^2.)

    We can choose any z_1, any even z_2 and generate a solution. Thus, there are infinitely many solutions.
    Last edited by alexmahone; September 7th 2011 at 03:49 AM.
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  8. #8
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    Re: yet another Diophantine equation

    Where did the y appear from?
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  9. #9
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    Re: yet another Diophantine equation

    Quote Originally Posted by procyon View Post
    Where did the y appear from?
    w=y-z, remember?
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  10. #10
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    Re: yet another Diophantine equation

    Quote Originally Posted by alexmahone View Post
    w=y-z, remember?
    I've got that bogged down in this that I'd forgotten, lol. Sorry.
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  11. #11
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    Re: yet another Diophantine equation

    Thank you so much to alexmahone and CaptainBlack for helping me with this.

    Here's the whole thing in case it's of use to someone else.
    (I've used p and q for the factors only because I find it easier to follow)

    (y-z)^2-x^2=2z^2

    so \{(y-z)+x\}\{(y-z)-x\}=2z^2

    let p and q be factors of z so that z=pq and 2z^2=2p^2q^2

    now \{(y-z)+x\}\{(y-z)-x\}=2p^2q^2


    let \ \ (y-z)+x=2p^2..........(1)

    and \ \ (y-z)-x=q^2..........(2)

    (1)+(2)\ \ \ \ \ y-z=\dfrac{2p^2+q^2}{2}\ \ \ \ and\ \ \ \ y=\dfrac{2p^2+2pq+q^2}{2}

    (1)-(2)\ \ \ \ \ x=\dfrac{2p^2-q^2}{2}

    and now for the checking...


    (y-z)^2-x^2=\left(\dfrac{2p^2+q^2}{2}\right)^2-\left(\dfrac{2p^2-q^2}{2}\right)^2=\dfrac{8p^2q^2}{4}=2p^2q^2

    and\ \ \ 2z^2=2(pq)^2=2p^2q^2

    Thanks again guys

    Pro
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  12. #12
    Grand Panjandrum
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    Re: yet another Diophantine equation

    Quote Originally Posted by alexmahone View Post
    Surely, you mean:

    w+x=2z_1^2 ---------- (1)

    w-x=z_2^2 ---------- (2)
    You are right that there is a typo, but what I actually meant was z_1z_2=z^2, that is that z_1 is a factor of z^2 and z_2=z^2/z_1.

    CB
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  13. #13
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    Re: yet another Diophantine equation

    Quote Originally Posted by CaptainBlack View Post
    You are right that there is a typo, but what I actually meant was z_1z_2=z^2, that is that z_1 is a factor of z^2 and z_2=z^2/z_1.

    CB
    In that case, I think your method is better than mine.

    There is no reason why w-x must be a perfect square or w+x must be twice a perfect square as long as (w+x)(w-x) turns out to be twice a perfect square.

    So, my method clearly leaves out some solutions.
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