# Simultaneous equations

x$$^{5}$$=5y$$^{3}$$-4z
y$$^{5}$$=5z$$^{3}$$-4x
z$$^{5}$$=5x$$^{3}$$-4y

We get five solutions (0,1,-1,2,-2) for x=y=z. But it's hard to do this when y$$\neq$$x$$\neq$$z.
Any ideas?

hotvette
Homework Helper
These are a set of non-linear simultaneous equations. I believe numerical methods are needed.

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?

hotvette
Homework Helper
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.

robphy
Homework Helper
Gold Member
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]

HallsofIvy
Homework Helper
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?

robphy
Homework Helper
Gold Member
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:
AlephZero
Homework Helper
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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