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Thread: Diophantine equation .

  1. #1
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    Diophantine equation .

    Hi everyone !
    Solve in N˛ this diophantine equation :
    $\displaystyle y^2=x^3-5x$
    Good luck !
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Tarask View Post
    Hi everyone !
    Solve in N˛ this diophantine equation :
    $\displaystyle y^2=x^3-5x$
    Good luck !
    Try to draw this function...

    You will see that you need to check only values of x between -2 and 2.
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  3. #3
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    Hi !
    I actually do have a solution using graphs , equation of tangents , but it is too long and of course not beautiful at all !
    I wanted a more "algebraic" solution , and i started this way :
    if x=0 then y=0
    suppose that x and y are different from 0
    If x<0
    $\displaystyle y^2=x^3-5x$ since y˛ is positif and x is negatif x˛-5 must be negatif and we get
    if x=-2 we get y˛=2 which is impossible , and if x=-1 we get y=2 or y=-2
    so (-1,2) and (-1,-2) are solutions.
    We now have to treat the case of x>0 , and here is where i got stuck !
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Try to draw this function...

    You will see that you need to check only values of x between -2 and 2.

    I can't see why: $\displaystyle (5,10)\,,\,(5,-10)$ are solutions, say...

    Tonio
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by tonio View Post
    I can't see why: $\displaystyle (5,10)\,,\,(5,-10)$ are solutions, say...

    Tonio

    Me too...
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  6. #6
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    Hi again !
    thanks for paying attention to my exercice !
    Here is what Wolfram gave me
    it also gave me that the only integer solutions are the ones i found , really weird !!!! I think we should use the gcd !
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  7. #7
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    For $\displaystyle x,y > 0$ :

    We have that $\displaystyle x\cdot (x^2 - 5)$ is a square, now of course $\displaystyle \text{gcd}(x,x^2-5)$ divides $\displaystyle x\cdot x + (-1)\cdot (x^2 - 5) = 5$ thus $\displaystyle \text{gcd}(x,x^2-5)$ is either 1 or 5 (in case that $\displaystyle 5|x$ )

    So we separate in the 2 cases:

    Case 1: $\displaystyle \text{gcd}(x,x^2-5) = 1$.

    Here it immediately follows that both $\displaystyle x$ and $\displaystyle x^2-5$ should be squares for their product to be a square.

    However, note that $\displaystyle x^2-5$ if $\displaystyle x$ is big enough (since the "distance" between 2 consecutive squares increases) it can't be an square since $\displaystyle x^2$ is already an square. How big is big enough? note that for $\displaystyle x>3$ it will be impossible, and in fact you can see we get no solutions for this case.

    Case 2: $\displaystyle \text{gcd}(x,x^2-5) = 5$.

    Here $\displaystyle x = 5\cdot k$ thus $\displaystyle y^2 = 5^2\cdot k \cdot (5\cdot k^2 - 1) $

    Thus $\displaystyle \left(\displaystyle\frac{y}{5}\right)^2 = k \cdot (5\cdot k^2 - 1)$ but now of course $\displaystyle k$ and $\displaystyle 5\cdot k^2 - 1$ are coprime and so they both must be squares.

    If $\displaystyle k=1$ we have a solution ( $\displaystyle x = 5$ ; $\displaystyle y=10$ ).

    $\displaystyle k = a^2$ then we want $\displaystyle 5\cdot a^4 -1 = b^2$

    mmmm... stuck.
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    An Idea...

    We need to prove that there are only solutions of the form: (a,2a) {and (a,-2a)}.

    ...and if we prove that we can make a conclusion that the only solution are: (0,0),(-1,2),(-1,-2),(5,10),(5,-10).
    {by solving: a^3-4a^2-5a=0}
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  9. #9
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    Quote Originally Posted by Also sprach Zarathustra View Post
    An Idea...

    We need to prove that there are only solutions of the form: (a,2a) {and (a,-2a)}.

    ...and if we prove that we can make a conclusion that the only solution are: (0,0),(-1,2),(-1,-2),(5,10),(5,-10).
    {by solving: a^3-4a^2-5a=0}
    Using parity may be ?
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    By showing that $\displaystyle \sqrt{2n(4n^2-5)}$ is not an integer we will conclude that $\displaystyle a$ is must to be odd.

    e.i:
    ...hence(after proving the above) $\displaystyle a$ is odd, say $\displaystyle a=2n+1$ it easy to show that $\displaystyle y$ as integer must be even:
    $\displaystyle y=\sqrt{(2n+1)^3-5(2n+1)}=\sqrt{4(2n+1)(n^2+n-1)}=2\sqrt{(2n+1)(n^2+n-1)}$
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  11. #11
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    Quote Originally Posted by Also sprach Zarathustra View Post
    By showing that $\displaystyle \sqrt{2n(4n^2-5)}$ is not an integer we will conclude that $\displaystyle a$ is must to be odd.
    Well if we consider the functions f such that $\displaystyle f(x)=8x^3-10x$ and g such that $\displaystyle g(x)=x^2$ we find that their graphs never meet if i'm not wrong of course ....
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  12. #12
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Tarask View Post
    Well if we consider the functions f such that $\displaystyle f(x)=8x^3-10x$ and g such that $\displaystyle g(x)=x^2$ we find that their graphs never meet if i'm not wrong of course ....
    Interesting!
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  13. #13
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Interesting!
    Thanks !
    By the way , as i said , i have a solution using graphs , it is similar to what we did !
    It's in french , if you are interested in it , i'll publish the file so that you can take a look at it
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  14. #14
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Tarask View Post
    Thanks !
    By the way , as i said , i have a solution using graphs , it is similar to what we did !
    It's in french , if you are interested in it , i'll publish the file so that you can take a look at it
    I will be glad to look at it.
    My French is like your Hebrew... Merci mon ami.
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  15. #15
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I will be glad to look at it.
    My French is like your Hebrew... Merci mon ami.
    Your french is good apparently , but for me , i don't know a single word of Hebrew
    Here is the file http://www.animath.fr/IMG/pdf/cours-arith1.pdf see page 68-.. (i hope that the mods and admins don't mind )
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