I can't find the solutions of this system of equations. I only need the "whole number" solutions. The solutions I'm searching for: $x=-1 \land y = 1 \land \lambda = \frac{1}{2e}$
\begin{cases} \begin{array} {rcl} e^{x+y-y^2}+2\lambda x \ = 0 \\ e^{x+y-y^2}(1-2y)+2\lambda y\ = 0 \\ x^2+y^2-2 \ = 0 \end{array} \end{cases}
Can someone give me some hints to solve this system? Thanks!
Given $(x,y,z)\in \mathbb{Z}$, $x^2+y^2-2=0$ is satified by $x = \pm 1,y = \pm 1$. Then \begin{eqnarray*} e^{x+y-y^{2}}+2\lambda x &=& 0 \end{eqnarray*}
can have the following possibilities,$e+2\lambda=0$, $\frac{1}{e}-2\lambda=0$, $\frac{1}{e}+2\lambda=0$ or $\frac{1}{e^{3}}-2\lambda=0$. This lead to $\lambda=\left\{-\frac{1}{2}e,\pm \frac{1}{2e},\frac{1}{2e^{3}}\right\}$
To satisfy,
\begin{eqnarray*} \left(1-2y\right)e^{x+y-y^{2}}+2\lambda y &=& 0 \end{eqnarray*}
the only possibility is $x=-1,y=1,\lambda=\frac{1}{2e}$.
