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

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== Solution ==
 
== Solution ==
{{solution}}
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Clearly <math>\boxed{(1,1,1)}</math> works. We then prove that no other solutions exist.
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<math>(1)</math> First, if <math>x>1</math>, then <math>y^2+y-1=\frac{1}{x}<1</math>, so <math>y^2+y-2<0</math> and thus <math>y\in(-2,1)</math>. Since <math>y</math> must be positive, then we have <math>0<y<1</math>.
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<math>(2)</math> Next, if <math>0<x<1</math>, then <math>y^2+y-1=\frac{1}{x}>1</math>, so <math>y^2+y-2>0</math> and thus <math>y\in(-\infty,-2)\cup(1,\infty)</math>. Since <math>y</math> must be positive, then we have <math>y>1</math>.
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Since <math>x=1</math> implies that <math>y=1</math> and <math>z=1</math> because they are all positive, this covers all possible values of <math>x</math>. Now, we use these properties on all variables.
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Assume that <math>0<x<1</math>. By <math>(2)</math>, we have <math>y>1</math>, and by <math>(1)</math>, we have <math>0<z<1</math>. Finally, by <math>(2)</math>, we have <math>x>1</math>, a contradiction, so no solutions exist in this case.
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Similarly, assume that <math>x>1</math>. By <math>(1)</math>, we have <math>0<y<1</math>, and by <math>(2)</math>, we have <math>z>1</math>. Finally, by <math>(1)</math>, we have <math>0<x<1</math>, a contradiction, so no solutions exist in this case either.
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As a result, this leaves <math>\boxed{(1,1,1)}</math> as our only solution.
  
 
== See also ==
 
== See also ==
 
[[OIM Problems and Solutions]]
 
[[OIM Problems and Solutions]]

Revision as of 17:20, 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.

See also

OIM Problems and Solutions