7.51 There are 1988 identical unit cubes, and they (all or part) are used to form three "squares" $A, B, C$ (i.e., three one-layer-high rectangular prisms with dimensions $a \times a \times 1, b \times b \times 1, c \times c \times 1$ where $a \leqslant b \leqslant c$). Now, place square $C$ on a flat surface, then place $B$ on top of $C$ so that each small block of $B$ is exactly on top of a small block of $C$, but the boundary of $B$ does not align with the sides of $C$. Finally, place $A$ on top of $B$ following the same principle, thus forming a "three-story building". When $a, b, c$ take what values, can the number of different "three-story buildings" that can be formed be maximized?
[Solution] Since when $a \leqslant b-2$, the number of different ways to place $A$ on $B$ without aligning the perimeters is $(b-a-1)^{2}$, this problem is equivalent to finding the maximum value of $(b-a-1)^{2} \cdot(c-b-1)^{2}$ under the condition
$$
1 \leqslant a \leqslant b-2 \leqslant c-4, a, b, c \in N, a^{2}+b^{2}+c^{2} \leqslant 1988
$$

Obviously, this maximum value is reached when $a=1$, so we only need to find the maximum value of $(b-2)(c-b-1)$ under the condition
$$
3 \leqslant b \leqslant c-2, b, c \in N, b^{2}+c^{2} \leqslant 1987
$$

For each fixed $c$, the sum of the two factors is constant, so the value of $b$ that minimizes the difference between the two factors $b-2$ and $c-b-1$ maximizes the product. Since $c^{2} \leqslant 1987-3^{2}=1978$, we know that $c \leqslant$ 44.
$$
\text { (1) } c=44, b^{2} \leqslant 1987-44^{2}=51, \quad 3 \leqslant b \leqslant 7 . \quad(43-b) \cdot(b
$$
-2 ) is maximized at $b=7$ with a value of 180;
(2) $c=43, b^{2} \leqslant 1987-43^{2}=138, \quad 3 \leqslant b \leqslant 11, \quad(42-b) \cdot(b$
-2) is maximized at $b=11$ with a value of 279;
(3) $c=42, b \leqslant 14,(41-b)(b-2)$ is maximized at $b=14$ with a value of 324;
(4) $c=41, b \leqslant 17,(40-b)(b-2)$ is maximized at $b=17$ with a value of 345;
(5) $c=40, b \leqslant 19,(39-b)(b-2)$ is maximized at $b=19$ with a value of 340;
(6) $c \leqslant 39,(b-2)(c-b-1) \leqslant\left(\frac{c-3}{2}\right)^{2} \leqslant 18^{2}=324$. It is evident that when $a=1, b=17, c=41$, the number of "three-story buildings" formed is the maximum (a total of $345^{2}$ ways).