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Simultaneous equations

  1. Sep 10, 2007 #1
    x[tex]^{5}[/tex]=5y[tex]^{3}[/tex]-4z
    y[tex]^{5}[/tex]=5z[tex]^{3}[/tex]-4x
    z[tex]^{5}[/tex]=5x[tex]^{3}[/tex]-4y

    We get five solutions (0,1,-1,2,-2) for x=y=z. But it's hard to do this when y[tex]\neq[/tex]x[tex]\neq[/tex]z.
    Any ideas?
     
  2. jcsd
  3. Sep 10, 2007 #2

    hotvette

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    These are a set of non-linear simultaneous equations. I believe numerical methods are needed.
     
  4. Sep 11, 2007 #3
    :|
    It wasn't hard to guess that it's non-linear. But have you got any idea?
     
  5. Sep 11, 2007 #4

    hotvette

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    Well, you can try Newton's method for solving simultaneous non-linear equations. Click the link in my signature for an example.

    Actually, I tried it for this problem for about 30 minutes and wasn't able to find any solution other than the ones you already know.
     
  6. Sep 11, 2007 #5

    robphy

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    can you take advantage of the symmetry in x,y,z?
    [you probably can see that each equation has odd powers in each of x,y, and z]
     
  7. Sep 11, 2007 #6

    HallsofIvy

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    That was why he first tried x= y= z which gives the solutions cited. The question was whether itis possible to get a solution with x, y, z different. Of if it is, then those same values permuted among x, y, z is also a solution.
     
  8. Sep 12, 2007 #7
    So, you suggest that all of acceptable values of x,y,z are only 0,1,-1,2,-2?
     
  9. Sep 12, 2007 #8

    robphy

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    Maple says so [for real solutions].
    I've seen a similar problem earlier... there must be a way to more methodically show that those are the only solutions for this apparently special set of symmetrical equations. If you assume that x=/=y, can it be shown that no solution with real z exists?
     
    Last edited: Sep 12, 2007
  10. Sep 12, 2007 #9

    AlephZero

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    You can show these are the only solution by a semi-graphical, semi-analytical argument. (which I guess you could make it more rigorous if you want to).

    The equations are
    x5=5y3-4z (1)
    y5=5z3-4x (2)
    z5=5x3-4y (3)

    Let z be some fixed value. Then the solutions of (1) are a set of curves, where dy/dx > 0 for every curve.

    Similarly the solutions of (2) are a set of curves where dy/dx < 0 everywhere.

    So for a given value of z, there is at most one solution (x,y,z) of equations (1) and (2), because the solution must be the interesection of a curve with positive slope (1) and a curve with negative slope (2). Plotting out the curves with a spreadsheet shows there is a unique solution to (1) and (2) for every value of z. The solutions lie on a smooth curve in 3D space.

    Plots of the (x,y) curves for (1) and (2), for z = -3, -2.5, -2, ... 2.5, 3, attached

    Using the same argument, the solutions to (2) and (3) also lie on a smooth curve.

    The solutions to all 3 equations are the intersections of these two curves.

    It's clear (by drawing pictures) the two curves diverge for large values of z and the only solutions are x = y = z = (-2, -1, 0, 1, 2)

    But this is a horrible argument, because it ignores the symmetry in the equations! Somebody do better, please!!
     

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