7.56 On the table, there are 1989 equally sized circular paper pieces placed without overlapping. How many different colors are needed at least to ensure that each paper piece can be painted a color so that any two touching circular paper pieces are painted different colors, regardless of their positions?
［Solution］Consider the case of 11 circular pieces of paper as shown in the figure. Suppose that the 6 circles on the left have been colored. Clearly, the 3 circles $A, B, E$ can only be colored with 1 or 3, and $A$ is one color, while $B$ and $E$ are the other color. If there are only 3 colors, then no matter what color $C$ and $D$ are painted, the requirements cannot be met. Therefore, to color these 11 circles and meet the requirements of the problem, at least 4 different colors are needed.

Below, we use mathematical induction to prove that as long as there are 4 different colors, the circles can be colored according to the requirements of the problem.

Assume that the conclusion holds when $n=k$. When $n=k+1$, consider the convex hull of the centers of these $k+1$ circles. Let $A$ be one of the vertices of this convex polygon. Clearly, the circle centered at $A$ can touch at most 3 other circles. According to the induction hypothesis, the remaining $k$ circles, excluding the one centered at $A$, can be colored with 4 colors as required. After coloring, the circles touching circle $A$ are at most 3, and of course, they can be painted with at most 3 different colors. Therefore, it is sufficient to paint circle $A$ with the 4th color.