Bài giải:

$\left| {{x}^{2}}-1 \right|+{{\left( {{y}^{2}}-3 \right)}^{2}}=2$

Vì (x ; y) nguyên ; y<0  và  $\left\{ \begin{align}& \left| {{x}^{2}}-1 \right|\ge 0 \\& {{\left( {{y}^{2}}-3 \right)}^{2}}\ge 0 \\ \end{align} \right.$  nên

$\left\{ \begin{align}& \left| {{x}^{2}}-1 \right|=1 \\ & {{\left( {{y}^{2}}-3\right)}^{2}}=1 \\ \end{align} \right.$  hay  $\left\{ \begin{align}& {{x}^{2}}-1=-1 \\  & {{y}^{2}}-3=1 \\ \end{align} \right.$

$\left\{ \begin{align}& x=0 \\ & y=-2 \\ \end{align} \right.$