1. Given that $x$ and $y$ are positive integers, and satisfy $x y + x + y = 71$, $x^{2} y + x y^{2} = 880$. Then $x^{2} + y^{2} =$ $\qquad$ .
$-、 1.146$.
Let $a=x+y, b=x y, x y+x+y=a+b=71$, $x^{2} y+x y^{2}=x y(x+y)=a b=880$.
Therefore, $a, b$ are the two roots of $t^{2}-71 t+880=0$.
Solving, we get $a=x+y, b=x y$ equal to 16 and 55, respectively.
If $x+y=55, x y=16$, there are clearly no positive integer solutions. Therefore, only $x+y=16, x y=55$.
Thus, $x^{2}+y^{2}=(x+y)^{2}-2 x y=146$.