559 (1) In the first quadrant of the Cartesian coordinate system, there are $(n+1)^{2}$ grid points with coordinates $(i, j)(i, j \in \mathbf{Z}, 1 \leqslant i, j \leqslant n)$. Find the number of squares that can be formed with these grid points as vertices (squares can be tilted);
(2) In space, does there exist a cube with vertices as grid points and all edges not parallel to the coordinate axes?
(1) First, the number of squares with vertices on the points of the edges of a $k \times k$ square is denoted as $k$. For example, as shown in Figure 1, for a $4 \times 4$ square, the squares with vertices on the 4 edges are the 4 squares with vertices at $A$, $B$, $C$, and $D$, so the number is 4.

Second, in an $n \times n$ square, there are $(n+1-k)^{2}$ squares of size $k \times k$. Each square with vertices at grid points must be on the edge of some $k \times k$ square. Therefore, the number of squares with vertices at grid points is
$$
\begin{array}{l}
\sum_{k=1}^{n} k(n+1-k)^{2} \\
= \sum_{k=1}^{n} k^{3}-2(n+1) \sum_{k=1}^{n} k^{2}+(n+1)^{2} \sum_{k=1}^{n} k \\
= \frac{n^{2}(n+2)^{2}}{4}- \\
2(n+1) \frac{n(n+1)(2 n+1)}{6}+ \\
(n+1)^{2} \frac{n(n+1)}{2} \\
= \frac{n(n+1)^{2}(n+2)}{12}.
\end{array}
$$

In summary, the number of squares with vertices at the grid points of an $n \times n$ square is $\frac{n(n+1)^{2}(n+2)}{12}$.
(2) In space, let one vertex of a cube be the origin, and the coordinates of the three adjacent vertices be
$$
\begin{array}{l}
(a, b, c),(d, e, f),(g, h, i) . \\
\text { Then } a^{2}+b^{2}+c^{2}=d^{2}+e^{2}+f^{2}=g^{2}+h^{2}+i^{2}, \\
a d+b e+c f=a g+b h+c i \\
=d g+e h+f i=0 .
\end{array}
$$

The integer solutions to the above indeterminate equations exist. For example,
$$
\begin{array}{l}
(a, b, c, d, e, f, g, h, i) \\
=(1,2,2,2,-2,1,2,1,-2)
\end{array}
$$

Clearly, these satisfy the conditions.