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

[phucghe]

Find all pairs $$(x,y) \in R$$ such that :
$$\frac{x^4-16}{8x}=\frac{y^4-1}{y}$$ and $$x^2-2xy+y^2=8$$

 

[Tac-Tics]

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?

 

[CRGreathouse]

Mathematica finds 8 complex solutions.

 

[phucghe]

I'am a elemantary pupil so I don't know about complex number
Could anyone give me a complete solution.

 

[Tac-Tics]

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.

 

[arildno]

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.

 

[phucghe]

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