# solve equations with high degree

this is my code,i want to solve this equations.can anyone give me some help about this,wheather mathematica can solve this.thanks a lot

eqn1:=(4 x + 12 x^3 + 12 x^5 + 4 x^7 - 4 y - 2 x^2 y + 6 x^4 y + 2 x^6 y -
2 x^8 y + 2 x y^2 + 12 x^3 y^2 + 10 x^5 y^2 - 12 y^3 -
12 x^2 y^3 + 2 x^4 y^3 - 2 x^8 y^3 - 6 x y^4 - 2 x^3 y^4 -
4 x^7 y^4 - 12 y^5 - 10 x^2 y^5 - 2 x^6 y^5 - 2 x y^6 +
2 x^5 y^6 - 4 y^7 + 4 x^4 y^7 + 2 x y^8 + 2 x^3 y^8 + 12 x^2 z +
24 x^4 z + 12 x^6 z + 24 x^3 y z + 24 x^5 y z - 12 y^2 z +
30 x^4 y^2 z + 8 x^6 y^2 z - 2 x^8 y^2 z - 24 x y^3 z +
8 x^5 y^3 z - 8 x^7 y^3 z - 24 y^4 z - 30 x^2 y^4 z -
8 x^6 y^4 z - 24 x y^5 z - 8 x^3 y^5 z - 12 y^6 z - 8 x^2 y^6 z +
8 x^4 y^6 z + 8 x^3 y^7 z + 2 x^2 y^8 z + 10 x z^2 + 32 x^3 z^2 +
26 x^5 z^2 + 4 x^7 z^2 - 10 y z^2 + 16 x^2 y z^2 + 53 x^4 y z^2 +
18 x^6 y z^2 - x^8 y z^2 - 16 x y^2 z^2 + 34 x^3 y^2 z^2 +
26 x^5 y^2 z^2 - 8 x^7 y^2 z^2 - 32 y^3 z^2 - 34 x^2 y^3 z^2 +
10 x^4 y^3 z^2 - 12 x^6 y^3 z^2 - 53 x y^4 z^2 - 10 x^3 y^4 z^2 -
4 x^5 y^4 z^2 - 26 y^5 z^2 - 26 x^2 y^5 z^2 + 4 x^4 y^5 z^2 -
18 x y^6 z^2 + 12 x^3 y^6 z^2 - 4 y^7 z^2 + 8 x^2 y^7 z^2 +
x y^8 z^2 + 20 x^2 z^3 + 32 x^4 z^3 + 12 x^6 z^3 + 44 x^3 y z^3 +
32 x^5 y z^3 - 4 x^7 y z^3 - 20 y^2 z^3 + 22 x^4 y^2 z^3 -
14 x^6 y^2 z^3 - 44 x y^3 z^3 - 10 x^5 y^3 z^3 - 32 y^4 z^3 -
22 x^2 y^4 z^3 - 32 x y^5 z^3 + 10 x^3 y^5 z^3 - 12 y^6 z^3 +
14 x^2 y^6 z^3 + 4 x y^7 z^3 + 8 x z^4 + 22 x^3 z^4 + 14 x^5 z^4 -
8 y z^4 + 13 x^2 y z^4 + 21 x^4 y z^4 - 6 x^6 y z^4 -
13 x y^2 z^4 + 6 x^3 y^2 z^4 - 12 x^5 y^2 z^4 - 22 y^3 z^4 -
6 x^2 y^3 z^4 - 4 x^4 y^3 z^4 - 21 x y^4 z^4 + 4 x^3 y^4 z^4 -
14 y^5 z^4 + 12 x^2 y^5 z^4 + 6 x y^6 z^4 + 8 x^2 z^5 +
8 x^4 z^5 + 4 x^3 y z^5 - 4 x^5 y z^5 - 8 y^2 z^5 -
4 x^4 y^2 z^5 - 4 x y^3 z^5 - 8 y^4 z^5 + 4 x^2 y^4 z^5 +
4 x y^5 z^5 + 2 x z^6 + 2 x^3 z^6 - 2 y z^6 - x^2 y z^6 -
x^4 y z^6 + x y^2 z^6 - 2 y^3 z^6 +
x y^4 z^6)/(8 (1 + x^2) (1 + y^2) (1 + x^2 + 2 x y + y^2) (1 +
x^2 + 2 x z + z^2) (1 + y^2 + 2 y z + z^2) (1 + x^2 + 2 x y +
y^2 + 2 x z + 2 y z + z^2));

eqn2:=((3 y)/(1 + y^2)^(5/2) + (3 (x + y))/(1 + (x + y)^2)^(5/2) + (
3 (y + z))/(1 + (y + z)^2)^(5/2) + (
3 (x + y + z))/(1 + (x + y + z)^2)^(
5/2)) (-((2 x^2)/(1 + x^2)^2) + 1/(1 + x^2) - (
2 x (x + y))/(1 + (x + y)^2)^2 + 1/(1 + (x + y)^2) - (
2 x (x + z))/(1 + (x + z)^2)^2 + 1/(1 + (x + z)^2) - (
2 x (x + y + z))/(1 + (x + y + z)^2)^2 + 1/(
1 + (x + y + z)^2)) + (-((3 x)/(1 + x^2)^(5/2)) - (
3 (x + y))/(1 + (x + y)^2)^(5/2) - (3 (x + z))/(1 + (x + z)^2)^(
5/2) - (3 (x + y + z))/(1 + (x + y + z)^2)^(
5/2)) (-((2 y^2)/(1 + y^2)^2) + 1/(1 + y^2) - (
2 y (x + y))/(1 + (x + y)^2)^2 + 1/(1 + (x + y)^2) - (
2 y (y + z))/(1 + (y + z)^2)^2 + 1/(1 + (y + z)^2) - (
2 y (x + y + z))/(1 + (x + y + z)^2)^2 + 1/(1 + (x + y + z)^2));
Reduce[eqn1==0&&eqn2==0,{x,y,z}]

• 1. You have 2 equations with 3 variables. Generally you need the third equation. 2. In principle, if you add a third equation you can try to solve them. Use then the Assignment(= ) instead of the DelayedAssignment (:= ). 3. You did not write, if you need an analytical or a numerical solution. In the latter case, what is the interval of variables, where you are going to look for solutions. 4. If you are looking for an analytical solutions, I doubt that it will work, but I would first try. I further doubt that the solution will be of any use, provided it will work. – Alexei Boulbitch Jan 7 '15 at 8:41
• I want to solve this equations and get a answer x==y and x==z or some similiars.other conditons x,y,z have the same sign x>0&&y>0&&z>0 or x<0&&y<0&&z<0.Please help me – star Jan 8 '15 at 7:55
• A) You did not answer my question N3. B) Your expressions are not equations Namely, the construction eqn:=x+y is an expression, while eqn=x+y==0 is the equation x+y==0 with the name "eqn". So you should fix to what your expressions are equal. Without this it is impossible to proceed further. I would also recommend to learn basics of the Mma syntax. – Alexei Boulbitch Jan 8 '15 at 8:13
• This is my fault,what you have said right. – star Jan 9 '15 at 7:00
• eqn2=1/8 (x/(1 + x^2) + x/(1 + (x + y)^2) + x/(1 + (x + z)^2) + x/( 1 + (x + y + z)^2))-1/8 (y/(1 + y^2) + y/(1 + (x + y)^2) + y/(1 + (y + z)^2) + y/( 1 + (x + y + z)^2)) Reduce[eqn1==0&&eqn2==0,{x,y,z}].Right?can you help me?? – star Jan 9 '15 at 7:05

You have a bad attitude not to answer the asked questions. For example, my question number 3. They are not asked without reason. I try to answer nevertheless. So first of all your equations:

eq1 = (4 x + 12 x^3 + 12 x^5 + 4 x^7 - 4 y - 2 x^2 y + 6 x^4 y +
2 x^6 y - 2 x^8 y + 2 x y^2 + 12 x^3 y^2 + 10 x^5 y^2 -
12 y^3 - 12 x^2 y^3 + 2 x^4 y^3 - 2 x^8 y^3 - 6 x y^4 -
2 x^3 y^4 - 4 x^7 y^4 - 12 y^5 - 10 x^2 y^5 - 2 x^6 y^5 -
2 x y^6 + 2 x^5 y^6 - 4 y^7 + 4 x^4 y^7 + 2 x y^8 + 2 x^3 y^8 +
12 x^2 z + 24 x^4 z + 12 x^6 z + 24 x^3 y z + 24 x^5 y z -
12 y^2 z + 30 x^4 y^2 z + 8 x^6 y^2 z - 2 x^8 y^2 z -
24 x y^3 z + 8 x^5 y^3 z - 8 x^7 y^3 z - 24 y^4 z -
30 x^2 y^4 z - 8 x^6 y^4 z - 24 x y^5 z - 8 x^3 y^5 z -
12 y^6 z - 8 x^2 y^6 z + 8 x^4 y^6 z + 8 x^3 y^7 z +
2 x^2 y^8 z + 10 x z^2 + 32 x^3 z^2 + 26 x^5 z^2 + 4 x^7 z^2 -
10 y z^2 + 16 x^2 y z^2 + 53 x^4 y z^2 + 18 x^6 y z^2 -
x^8 y z^2 - 16 x y^2 z^2 + 34 x^3 y^2 z^2 + 26 x^5 y^2 z^2 -
8 x^7 y^2 z^2 - 32 y^3 z^2 - 34 x^2 y^3 z^2 + 10 x^4 y^3 z^2 -
12 x^6 y^3 z^2 - 53 x y^4 z^2 - 10 x^3 y^4 z^2 -
4 x^5 y^4 z^2 - 26 y^5 z^2 - 26 x^2 y^5 z^2 + 4 x^4 y^5 z^2 -
18 x y^6 z^2 + 12 x^3 y^6 z^2 - 4 y^7 z^2 + 8 x^2 y^7 z^2 +
x y^8 z^2 + 20 x^2 z^3 + 32 x^4 z^3 + 12 x^6 z^3 +
44 x^3 y z^3 + 32 x^5 y z^3 - 4 x^7 y z^3 - 20 y^2 z^3 +
22 x^4 y^2 z^3 - 14 x^6 y^2 z^3 - 44 x y^3 z^3 -
10 x^5 y^3 z^3 - 32 y^4 z^3 - 22 x^2 y^4 z^3 - 32 x y^5 z^3 +
10 x^3 y^5 z^3 - 12 y^6 z^3 + 14 x^2 y^6 z^3 + 4 x y^7 z^3 +
8 x z^4 + 22 x^3 z^4 + 14 x^5 z^4 - 8 y z^4 + 13 x^2 y z^4 +
21 x^4 y z^4 - 6 x^6 y z^4 - 13 x y^2 z^4 + 6 x^3 y^2 z^4 -
12 x^5 y^2 z^4 - 22 y^3 z^4 - 6 x^2 y^3 z^4 - 4 x^4 y^3 z^4 -
21 x y^4 z^4 + 4 x^3 y^4 z^4 - 14 y^5 z^4 + 12 x^2 y^5 z^4 +
6 x y^6 z^4 + 8 x^2 z^5 + 8 x^4 z^5 + 4 x^3 y z^5 -
4 x^5 y z^5 - 8 y^2 z^5 - 4 x^4 y^2 z^5 - 4 x y^3 z^5 -
8 y^4 z^5 + 4 x^2 y^4 z^5 + 4 x y^5 z^5 + 2 x z^6 + 2 x^3 z^6 -
2 y z^6 - x^2 y z^6 - x^4 y z^6 + x y^2 z^6 - 2 y^3 z^6 +
x y^4 z^6)/(8 (1 + x^2) (1 + y^2) (1 + x^2 + 2 x y + y^2) (1 +
x^2 + 2 x z + z^2) (1 + y^2 + 2 y z + z^2) (1 + x^2 + 2 x y +
y^2 + 2 x z + 2 y z + z^2)) == 0;
eq2 = ((3 y)/(1 + y^2)^(5/2) + (3 (x + y))/(1 + (x + y)^2)^(5/
2) + (3 (y + z))/(1 + (y + z)^2)^(5/
2) + (3 (x + y + z))/(1 + (x + y + z)^2)^(5/
2)) (-((2 x^2)/(1 + x^2)^2) +
1/(1 + x^2) - (2 x (x + y))/(1 + (x + y)^2)^2 +
1/(1 + (x + y)^2) - (2 x (x + z))/(1 + (x + z)^2)^2 +
1/(1 + (x + z)^2) - (2 x (x + y + z))/(1 + (x + y + z)^2)^2 +
1/(1 + (x + y + z)^2)) + (-((3 x)/(1 + x^2)^(5/2)) - (3 (x +
y))/(1 + (x + y)^2)^(5/
2) - (3 (x + z))/(1 + (x + z)^2)^(5/
2) - (3 (x + y + z))/(1 + (x + y + z)^2)^(5/
2)) (-((2 y^2)/(1 + y^2)^2) +
1/(1 + y^2) - (2 y (x + y))/(1 + (x + y)^2)^2 +
1/(1 + (x + y)^2) - (2 y (y + z))/(1 + (y + z)^2)^2 +
1/(1 + (y + z)^2) - (2 y (x + y + z))/(1 + (x + y + z)^2)^2 +
1/(1 + (x + y + z)^2)) == 0;


The use of Reduce with such equations is questionable. It fails with less complex equations. Solve[{eq1,eq2},{x,y}]might be better. However, in my case it worked non-stop over 5 minutes, after which I stopped it. You may try to start it and let working longer, if you like. It may finally solve the problem. However, this long calculation may be an indication that probably Mma can find no exact solution. Once more, it is not a proof, but an indication. Further, I do not know, what for are you looking for this solution, but in my experience exact solutions of such huge equation, if exist, are often useless, since they are typically enormously huge, and one cannot use them, say, to insert into another formula, or transform , or whatever. But you might choose to try nevertheless.

Further, if you would like to look for a numerical solution of your system, you should expect that it has several solutions and correspondingly, the numeric process should be expected to have several basins of attractors. I will only show it in one case by arbitrarily chosen initial points. The solution may be obtained by using the function FindRoot[{eq1,eq2},{{x,x0},{y,y0}}] at any fixed z. Then one can construct a list with the structure {...,{z,x,y}...} and do what? This depends upon your aims, of which I have no idea. So to construct the list one can do as follows:

lst = Table[{z,
FindRoot[{eq1, eq2}, {{x, 1}, {y, 10}}] /. {x -> a_,
y -> b_} -> {a, b}}, {z, 0, 1, 0.05}] /. {x_, {y_, z_}} -> {x,
y, z}


Now one can e.g., visualize it:

 Show[{
Graphics3D[{Blue, Thickness[0.005], Line[#]}, Axes -> True,
AxesLabel -> {Style["z", 16], Style["x", 16], Style["y", 16]}] &[
Partition[lst, 2, 1]],
Graphics3D[{Red, PointSize[0.015], Point[#]}] &[lst]
}]


which should return the following 3D image:

Here the points (red) which are nearest along z axis are connected by lines to guide the eye.

Be careful with this solution, since there has been a warning message that at least for one of the points the solution is questionable. If you go this way, try to play with (a) initial points and (b) method (check Menu/Help/Wolfram Documentation/FindRoot/Options/Method) to completely get rid of any warnings. Please be aware that I do not try to solve your problem, but to show the way I would treat it. Have fun!

• I'm so sorry to you,I want an analytical solutions.My first goal is to proof x==y,x==z,but I plot the use RegionPlot3D[eqn1==0&&eqn2==0,{x,-10,10},{y,-10,10},{z,-10,10},PlotPoints->100],Then i get x==y,So the goal of me if to proof x==y.Andditionally,the qualitily of equations is the two what i have said.Maybe,When you say this equations,you can know what i purpose – star Jan 10 '15 at 0:33
• eqn1=1/8 (x/(1 + x^2) + x/(1 + (x + y)^2) + x/(1 + (x + z)^2) + x/( 1 + (x + y + z)^2))-1/8 (y/(1 + y^2) + y/(1 + (x + y)^2) + y/(1 + (y + z)^2) + y/( 1 + (x + y + z)^2)) – star Jan 10 '15 at 0:34
• The eqn2 is the same with your text.This isn't my bad attitude,because i can't write it,it warning too larger.So i hope you can understand me.Please give me a help.Thanks a lot.And I will undertand the answer you have done.Thanks. – star Jan 10 '15 at 0:39