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 September 17th, 2017, 04:50 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 04:52 AM.
 September 17th, 2017, 05:27 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,655 Thanks: 841 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! Thanks from greg1313
 September 17th, 2017, 05:57 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, 05:58 AM   #4
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Quote:
 Originally Posted by romsek They are also totally symmetric so the solution must be $x=y=z$
Why do you say that? As another example, there are (real) solutions to the system $x^2 + y = 1$, $y^2 + x = 1$ such that $x \neq y$.

September 17th, 2017, 06:49 AM   #5
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 Originally Posted by cjem Why do you say that? As another example, there are (real) solutions to the system $x^2 + y = 1$, $y^2 + x = 1$ such that $x \neq y$.
You have a point but I believe only if some of those solutions set 1 or more of the variables to 0, thus essentially destroying the symmetry.

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, 07:16 AM   #6
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Quote:
 Originally Posted by romsek You have a point but I believe only if some of those solutions set 1 or more of the variables to 0, thus essentially destroying the symmetry. The solution that doesn't involve 0 to the system you note above does indeed have $x=y= \dfrac 1 2 (-1+\sqrt{5})$
Then consider the system $x^2 + y = 4$, $y^2 + x = 4$. There are real solutions with $x$ and $y$ distinct and non-zero.

 September 17th, 2017, 08:39 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,072 Thanks: 722 Are you ok cjem? http://www.wolframalpha.com/input/?i...,x%2By%5E2%3D4 Last edited by Denis; September 17th, 2017 at 08:44 AM.
September 17th, 2017, 08:59 AM   #8
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Quote:
 Originally Posted by Denis Are you ok cjem? Wolfram|Alpha: Computational Knowledge Engine
The top two solutions listed have $x$ and $y$ distinct and non-zero. This shows that what Romsek suggested in post #5 (that any solution to a "symmetric" set of equations with the variables not all equal must have one of the variables set to zero) is false.

Last edited by cjem; September 17th, 2017 at 09:57 AM.

 September 17th, 2017, 09:07 AM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,072 Thanks: 722 So? Same as (for example) 2+3=5 and 3+2=5
 September 17th, 2017, 09:15 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,072 Thanks: 722 For keerist's sake....

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