2016 OIM Problems/Problem 2

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

Page

Toolbox

Search

2016 OIM Problems/Problem 2

Problem

Solution

See also