1. Find all integers $x$ and $y$ that satisfy the equation $\left|x^{2}+2 x y-3 y^{2}\right|=1$.
II/1. Since $x^{2}+2 x y-3 y^{2}=(x+y)^{2}-4 y^{2}=(x-y)(x+3 y)$, we must consider four possibilities. For $x-y=1$ and $x+3 y=1$, we get the solution $x=1, y=0$. For $x-y=-1$ and $x+3 y=-1$, we get the solution $x=-1, y=0$. For $x-y=1$ and $x+3 y=-1$, we get the solution $x=1 / 2, y=-1 / 2$, but these numbers are not integers. For $x-y=-1$ and $x+3 y=1$, we get the solution $x=-1 / 2$, $y=1 / 2$, but these numbers are also not integers. Therefore, the equation has solutions $x=1$ and $y=0$ as well as $x=-1$ and $y=0$.

Factorization: $x^{2}+2 x y-3 y^{2}=(x-y)(x+3 y)$ : 3 points. Analysis of each of the 4 cases: each case 1 point. If the contestant only writes down a solution without proving uniqueness: 1 point for each solution.