## Problem A1

Find all real solutions to: $\mathrm{x}+\mathrm{y}-\mathrm{z}=-1 ; \mathrm{x}^{2}-\mathrm{y}^{2}+\mathrm{z}^{2}=1,-\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}=-1$.


## Solution

Answer: $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,-1,1)$ or $(-1,-1,-1)$.

From the first equation $x=z-y-1$. Substituting in the second equation: $2 z^{2}-2 y z+2 y-2 z=$ 0 , so $(\mathrm{z}-1)(\mathrm{z}-\mathrm{y})=0$. Hence $\mathrm{z}=1$ or $\mathrm{y}=\mathrm{z}$. If $\mathrm{z}=1$, then from the first equation $\mathrm{x}+\mathrm{y}=0$, and hence from the last equation, $\mathrm{x}=1, \mathrm{y}=-1$. If $\mathrm{y}=\mathrm{z}$, then $\mathrm{x}=-1$, and hence from the last

equation $\mathrm{y}=\mathrm{z}=-1$.

