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September 17th, 2017, 03:50 AM  #1 
Newbie Joined: Sep 2017 From: NY Posts: 6 Thanks: 0  interesting sets of equations
in real numbers sets of equations \begin{align*} x^{2}y+2=x+2yz \end{align*} \begin{align*} y^{2}z+2=y+2zx \end{align*} \begin{align*} z^{2}x+2=z+2xy \end{align*} Last edited by xps32; September 17th, 2017 at 03:52 AM. 
September 17th, 2017, 04:27 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757 
wow those are interesting! They are also totally symmetric so the solution must be $x=y=z$ Given this $x^3 + 2 = x+2x^2$ $x^3  2x^2  x + 2 = 0$ with real solutions $x = y = z = 1$ $x = y = z = 1$ $x = y = z = 2$ Interestingly enough, looking at the complex solutions, they do not exhibit the symmetry of the real solutions. I suppose they are distributed about the complex plane in some symmetric fashion. Further investigation is called for. Sounds like a job for Denis! 
September 17th, 2017, 04:57 AM  #3 
Newbie Joined: Sep 2017 From: NY Posts: 6 Thanks: 0 
show symmetry Sure enough to write so what you have written. I already knew 
September 17th, 2017, 04:58 AM  #4 
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  
September 17th, 2017, 05:49 AM  #5  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757  Quote:
The solution that doesn't involve 0 to the system you note above does indeed have $x=y= \dfrac 1 2 (1+\sqrt{5})$  
September 17th, 2017, 06:16 AM  #6 
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  Then consider the system $x^2 + y = 4$, $y^2 + x = 4$. There are real solutions with $x$ and $y$ distinct and nonzero.

September 17th, 2017, 07:39 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,688 Thanks: 701  Last edited by Denis; September 17th, 2017 at 07:44 AM. 
September 17th, 2017, 07:59 AM  #8  
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  Quote: Last edited by cjem; September 17th, 2017 at 08:57 AM.  
September 17th, 2017, 08:07 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,688 Thanks: 701 
So? Same as (for example) 2+3=5 and 3+2=5 
September 17th, 2017, 08:15 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,688 Thanks: 701 
For keerist's sake....


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