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September 17th, 2017, 04:50 AM   #1
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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.
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September 17th, 2017, 05:27 AM   #2
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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!
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September 17th, 2017, 05:57 AM   #3
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show symmetry
Sure enough to write so
what you have written. I already knew
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September 17th, 2017, 05:58 AM   #4
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Quote:
Originally Posted by romsek View Post
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 $x \neq y$.
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September 17th, 2017, 06:49 AM   #5
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Quote:
Originally Posted by cjem View Post
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 $x=y= \dfrac 1 2 (-1+\sqrt{5})$
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September 17th, 2017, 07:16 AM   #6
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Quote:
Originally Posted by romsek View Post
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.
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September 17th, 2017, 08:39 AM   #7
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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.
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September 17th, 2017, 08:59 AM   #8
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Quote:
Originally Posted by Denis View Post
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.
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September 17th, 2017, 09:07 AM   #9
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So?
Same as (for example) 2+3=5 and 3+2=5
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September 17th, 2017, 09:15 AM   #10
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For keerist's sake....
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