17. Let $x, y, z$ be integers, and
$$
x+y+z=3, x^{3}+y^{3}+z^{3}=3 \text {. }
$$

Then $x^{2}+y^{2}+z^{2}=$ $\qquad$ .
17.3 or 57.

Substitute $z=3-x-y$ into $x^{3}+y^{3}+z^{3}=3$, we get $x y=3(x+y)-9+\frac{8}{x+y}$.
Since $x, y$ are integers, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x + y = 1 , } \\
{ x y = 2 ; }
\end{array} \quad \left\{\begin{array}{l}
x+y=4, \\
x y=5 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x + y = 2 , } \\
{ x y = 1 ; }
\end{array} \quad \left\{\begin{array}{l}
x+y=8, \\
x y=16 .
\end{array},\right.\right.
\end{array}
$$

The first two systems of equations have no solutions.
From the last two systems of equations, we get
$$
x=y=z=1 ; x=y=4, z=-5 \text {. }
$$

Therefore, $x^{2}+y^{2}+z^{2}=3$ or 57.