17. Let $x, y, z$ be integers, and $x+y+z=3, x^{3}+y^{3}+z^{3}=3$, then $x^{2}+y^{2}+z^{2}=$
3 or 57.
Solution: Substituting $z=3-x-y$ into $x^{3}+y^{3}+z^{3}=3$ yields $x y=3(x+y)-9+\frac{8}{x+y}$. Since $x, y$ are integers, we have $\left\{\begin{array}{l}x+y=1 \\ x y=2\end{array},\left\{\begin{array}{l}x+y=4 \\ x y=5\end{array},\left\{\begin{array}{l}x+y=2 \\ x y=1\end{array},\left\{\begin{array}{l}x+y=8 \\ x y=16\end{array}\right.\right.\right.\right.$, 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.