17 Let $x, y, z$ be integers, and $x+y+z=3, x^{3}+y^{3}+z^{3}=3$, then $x^{2}+y^{2}+z^{2}=$ $\qquad$ .
173 or 57 Hint: Substituting $z=3-x-y$ 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
$$
\left\{\begin{array} { l } 
{ x + y = 1 , } \\
{ x y = 2 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x + y = 4 , } \\
{ x y = 5 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x + y = 2 , } \\
{ x y = 1 , }
\end{array} \quad \left\{\begin{array}{l}
x+y=8, \\
x y=16,
\end{array}\right.\right.\right.\right.
$$

The first two systems of equations have no solutions; the last two systems of equations solve to $x=y=z=1 ; x=y=4$, $z=-5$. Therefore, $x^{2}+y^{2}+z^{2}=3$ or 57.