What are the solutions to a system of equations involving x, y, and z?

  • Thread starter Thread starter phucghe
  • Start date Start date
  • Tags Tags
    System
AI Thread Summary
The discussion revolves around finding pairs (x,y) in R that satisfy the equations \(\frac{x^4-16}{8x}=\frac{y^4-1}{y}\) and \(x^2-2xy+y^2=8\). Participants note that Mathematica identifies eight complex solutions, but the challenge lies in proving these are the only solutions and finding real pairs. Suggestions include rewriting the equations to simplify the analysis, but concerns about continuity and the behavior of the function \(f(x)=x^3-\frac{1}{x}\) at x=0 complicate matters. The conversation emphasizes the difficulty of deriving intuitive solutions from the given equations. Overall, the problem remains unsolved, highlighting the complexity of the equations involved.
phucghe
Messages
4
Reaction score
0
Find all pairs (x,y) \in R such that :
\frac{x^4-16}{8x}=\frac{y^4-1}{y} and x^2-2xy+y^2=8
 
Mathematics news on Phys.org
phucghe said:
Find all pairs (x,y) \in R such that :
\frac{x^4-16}{8x}=\frac{y^4-1}{y} and x^2-2xy+y^2=8

1) You should write "pairs (x,y) \in R^2" or "x, y \in R".

2) What form do you need the answer in? Looking at those 4th powers and mixed terms, I'm guessing that there might not be a simple or intuitive solution for this.

3) What work have you done on the problem so far?
 
Mathematica finds 8 complex solutions.
 
I'am a elemantary pupil so I don't know about complex number
Could anyone give me a complete solution.
 
CRGreathouse said:
Mathematica finds 8 complex solutions.

Mathematica > my rough analysis of the problem.

If there's only 8 solutions, you can probably find them all by trial and error. To prove that there are exactly 8 solutions, and you have accounted for them all probably requires you to do a little arguing.
 
phucghe said:
Find all pairs (x,y) \in R such that :
\frac{x^4-16}{8x}=\frac{y^4-1}{y} and x^2-2xy+y^2=8
Rewrite your equations as:
\frac{(x-2)(x+2)(x^{2}+4)}{8x}=\frac{(y-1)(y+1)(y^{2}+1)}{y},(x-y)^{2}=8
this ought to help a bit.
 
thanks but it seems to be not necessary for this problem.
Though by putting x=2z I had : \frac{z^4-1}{z}=\frac{y^4-1}{y},cossidering the function f(x)=x^3-\frac{1}{x} and its monotonousness ,there are still some troubles for example f(x) is not continuous at x=0
 

Similar threads

Back
Top