# How to solve this system of equations?

1. Sep 29, 2012

### lape99

I don't know if I'm posting in the wright section, but i need to solve this system of equations:

{5x^4 - 4x + 5y = 0
5y^4 + 5x - 4y = 0}

I know one answer is (0;0), but i don't know how to show it, also there is another point.
Maybe someone can help me?

2. Sep 29, 2012

### arildno

If you subtract the equations from each other, you get the following:

5x^4-5y^4-9(x-y)=0

Now, you have:
x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3), and thus, you may factorize the above equation as:
(x-y)*(5(x^3+x^2y+xy^2+y^3)-9)=0

3. Sep 29, 2012

### lape99

Thank you very much :)

4. Oct 13, 2012

### Ray Vickson

One can use Groebner basis methods to get at all the solutions. Using the Groebner package in Maple 11, I get:
sys:=[5*x^4-4*x+5*y,5*y^4+5*x-4*y]:lprint(sys); <== input
[5*x^4-4*x+5*y, 5*y^4+5*x-4*y]
with(Groebner): <=== load the Groebner package
B:=Basis(sys,plex(x,y)):lprint(B);
[225*y+756*y^4-1280*y^7+2400*y^10-2000*y^13+625*y^16, 5*y^4+5*x-4*y]

This B is the output. The two original equations are equivalent to setting the two components of B to zero (although, of course, there may be some extraneous roots, so those must be checked in the original system). Note that the first component of B is a polynomial in y alone, so we can find roots of the two-equation system by first solving the polynomial in y, then solving for x by setting the second expression in B to zero.

f1y:=B[1]:lprint(f1y);
225*y+756*y^4-1280*y^7+2400*y^10-2000*y^13+625*y^16 <--- set to zero

Let's try to factor the polynomial.

F:=factor(f1y):lprint(F);
y*(5*y^3+1)*(125*y^12-425*y^9+565*y^6-369*y^3+225) <=== the factors

So, y = 0 or 5*y^3 + 1 = 0, or
125*y^12 -425*y^9 + 565*y^6 -360*y^3 + 225 = 0, which is a 4th degree polynomial in z = y^3.

After solving for y, we get x from 5*x = 4*y - 5*y^4.

Note: doing anything like this by hand would take months or years of work and require hundreds of pages of algebraic work. Use of a computer algebra system is essential.

RGV