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Difference between revisions of "2016 OIM Problems/Problem 2"

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As a result, this leaves <math>\boxed{(1,1,1)}</math> as our only solution.
 
As a result, this leaves <math>\boxed{(1,1,1)}</math> as our only solution.
 +
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
[[OIM Problems and Solutions]]
 
[[OIM Problems and Solutions]]

Latest revision as of 17:40, 23 March 2025

Problem

Find all positive real solutions of the system of equations: \[x=\frac{1}{y^2+y-1},\; y=\frac{1}{z^2+z-1},\; z=\frac{1}{x^2+x-1}\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Clearly $\boxed{(1,1,1)}$ works. We then prove that no other solutions exist.

$(1)$ First, if $x>1$, then $y^2+y-1=\frac{1}{x}<1$, so $y^2+y-2<0$ and thus $y\in(-2,1)$. Since $y$ must be positive, then we have $0<y<1$.

$(2)$ Next, if $0<x<1$, then $y^2+y-1=\frac{1}{x}>1$, so $y^2+y-2>0$ and thus $y\in(-\infty,-2)\cup(1,\infty)$. Since $y$ must be positive, then we have $y>1$.

Since $x=1$ implies that $y=1$ and $z=1$ because they are all positive, this covers all possible values of $x$. Now, we use these properties on all variables.

Assume that $0<x<1$. By $(2)$, we have $y>1$, and by $(1)$, we have $0<z<1$. Finally, by $(2)$, we have $x>1$, a contradiction, so no solutions exist in this case.

Similarly, assume that $x>1$. By $(1)$, we have $0<y<1$, and by $(2)$, we have $z>1$. Finally, by $(1)$, we have $0<x<1$, a contradiction, so no solutions exist in this case either.

As a result, this leaves $\boxed{(1,1,1)}$ as our only solution.

~ eevee9406

See also

OIM Problems and Solutions

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