Can Graphing Reveal All Solutions to Complex Polynomial Systems?

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The discussion focuses on solving a system of polynomial equations: 5x^4 - 4x + 5y = 0 and 5y^4 + 5x - 4y = 0, with (0,0) identified as one solution. Participants suggest using Groebner basis methods for finding all solutions, highlighting the efficiency of computer algebra systems like Maple 11. The Groebner basis output reveals a polynomial in y that can be factored to find additional roots. The process emphasizes that manual calculations for such complex systems would be impractical, underscoring the necessity of computational tools. Overall, graphing and algebraic techniques can effectively reveal all solutions to complex polynomial systems.
lape99
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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?
 
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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
 
Thank you very much :)
 
lape99 said:
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?

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
 
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