Can Symmetry Help Us Solve Simultaneous Equations with Different Variables?

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Discussion Overview

The discussion revolves around solving a set of non-linear simultaneous equations involving three variables (x, y, z) and explores the potential role of symmetry in finding solutions. Participants consider numerical methods, graphical approaches, and the implications of symmetry in the equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants identify the equations as non-linear and suggest that numerical methods may be necessary for finding solutions.
  • One participant proposes using Newton's method but notes difficulty in finding solutions beyond the trivial case where x = y = z.
  • Another participant questions whether the symmetry in the equations can be leveraged to find solutions where x, y, and z are different.
  • It is suggested that if a solution exists with x ≠ y ≠ z, then permutations of those values should also be solutions.
  • One participant mentions that Maple indicates the only real solutions are 0, 1, -1, 2, and -2, and questions if it can be shown that no solutions exist for x ≠ y.
  • A later reply proposes a semi-graphical, semi-analytical argument to show that for fixed values of z, there is at most one solution for the equations, based on the slopes of the curves derived from the equations.
  • However, this argument is critiqued for neglecting the symmetry in the equations, prompting a call for a better approach.

Areas of Agreement / Disagreement

Participants express differing views on the methods to solve the equations and the implications of symmetry. There is no consensus on whether solutions exist for x, y, and z being different, and the discussion remains unresolved regarding the best approach to take.

Contextual Notes

Limitations include the potential oversight of symmetry in the equations and the reliance on graphical methods that may not fully capture the complexity of the solutions.

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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?
 
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These are a set of non-linear simultaneous equations. I believe numerical methods are needed.
 
hotvette said:
These are a set of non-linear simultaneous equations. I believe numerical methods are needed.

:|
It wasn't hard to guess that it's non-linear. But have you got any idea?
 
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.
 
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]
 
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.
 
So, you suggest that all of acceptable values of x,y,z are only 0,1,-1,2,-2?
 
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:
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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