2. Given that $x$ and $y$ are positive integers, and satisfy $2 x^{2}+3 y^{2}$ $=4 x^{2} y^{2}+1$. Then $x^{2}+y^{2}=$ $\qquad$ .
2. From the given, we have $4 x^{2} y^{2}-2 x^{2}-3 y^{2}=-1$. Multiplying both sides by 4 and rearranging, we get
$$
\left(4 x^{2}-3\right)\left(4 y^{2}-2\right)=2=1 \times 2 \text {. }
$$
Solving, we get $\left\{\begin{array}{l}x^{2}=1, \\ y^{2}=1\end{array}\right.$ or $\left\{\begin{array}{l}x^{2}=1.25, \\ y^{2}=0.75 .\end{array}\right.$ (discard)