Find all triplets $(x,y,z)\in\mathbb{R}^3$ such that \begin{align*} x^2-yz &= |y-z|+1, \\ y^2-zx &= |z-x|+1, \\ z^2-xy &= |x-y|+1. \end{align*}
Problem
Source: Czech and Slovak Olympiad 2019, National Round, Problem 1
Tags: algebra, national olympiad, system of equations
Tintarn
08.06.2019 12:27
Interchanging $x,y,z$ by $(-x,-y,-z)$ we may w.l.o.g. assume that $x+y+z \ge 0$.
This is symmetric in $x,y,z$ so we may w.l.o.g. suppose that $x \ge y \ge z$ so that the system reads as
\begin{align*}
x^2-yz&=y-z+1\\
y^2-zx&=x-z+1\\
z^2-xy &=x-y+1
\end{align*}Subtracting the second from the first equation we find that $(y-z)(x+y+z-1)=0$ so that $y=z$ or $x+y+z=1$.
Subtracting the third from the second equation we find that $(x-y)(x+y+z+1)=0$ so that $x=y$ or $x+y+z=-1$.
Since clearly $x=y=z$ is not a solution, we must have $x=y$ and $x+y+z=1$.
Hence $x=-2y-1$ and the first equation is $x^2-y^2=1$ so that we find $(2y+1)^2-y^2=1$ and hence $y=0$ or $y=-\frac{4}{3}$.
However $y=0$ implies $x=-1<y$ which contradicts our assumption so the only solution in this case is $x=\frac{5}{3}, y=z=-\frac{4}{3}$.
Of course this gives rise to $6$ solutions by permuting the variables and introducing signs again.
Math5000
05.12.2019 15:37
Tintarn wrote:
Interchanging $x,y,z$ by $(-x,-y,-z)$ we may w.l.o.g. assume that $x+y+z \ge 0$.
This is symmetric in $x,y,z$ so we may w.l.o.g. suppose that $x \ge y \ge z$ so that the system reads as
\begin{align*}
x^2-yz&=y-z+1\\
y^2-zx&=x-z+1\\
z^2-xy &=x-y+1
\end{align*}Subtracting the second from the first equation we find that $(y-z)(x+y+z-1)=0$ so that $y=z$ or $x+y+z=1$.
Subtracting the third from the second equation we find that $(x-y)(x+y+z+1)=0$ so that $x=y$ or $x+y+z=-1$.
Since clearly $x=y=z$ is not a solution, we must have $x=y$ and $x+y+z=1$.
Hence $x=-2y-1$ and the first equation is $x^2-y^2=1$ so that we find $(2y+1)^2-y^2=1$ and hence $y=0$ or $y=-\frac{4}{3}$.
However $y=0$ implies $x=-1<y$ which contradicts our assumption so the only solution in this case is $x=\frac{5}{3}, y=z=-\frac{4}{3}$.
Of course this gives rise to $6$ solutions by permuting the variables and introducing signs again.
wonder if you renumbered your equations when writing up your answer since right at the start you say "Subtracting the second from the first equation" when it's actually "Subtracting the third from the second equation". Later in the solution you have y=x but then seem to switch to y=z
Moubinool
05.12.2019 18:49
x,y,z -5/3 , 4/3 , 4/3 -4/3,-4/3,5/3 and other permutation
hikmatullo
21.04.2022 08:26
well done