An $n\times n$ square is divided into $n^2$ unit cells. Is it possible to cover this square with some layers of 4-cell figures of the following shape [img]https://cdn.artofproblemsolving.com/attachments/5/7/d42a8011ec4c5c91c337296d8033d412fade5c.png[/img](i.e. each cell of the square must be covered with the same number of these figures) if
a) $n=6$? 
b) $n=7$?

(The sides of each figure must coincide with the sides of the cells; the figures may be rotated and turned over, but none of them can go beyond the bounds of the square.)
To determine if it is possible to cover an \( n \times n \) square with the given 4-cell figures, we need to consider the constraints and properties of the figures and the square.

### Part (a): \( n = 6 \)

1. **Coloring the Square:**
   Color the \( 6 \times 6 \) square in a chessboard pattern with alternating black and white cells. This means there will be 18 black cells and 18 white cells.

2. **Analyzing the 4-cell Figure:**
   Each 4-cell figure covers 3 cells of one color and 1 cell of the other color. Let \( a \) be the number of figures that cover 3 black cells and 1 white cell, and \( b \) be the number of figures that cover 3 white cells and 1 black cell.

3. **Setting Up Equations:**
   The total number of cells covered by the figures is:
   \[
   4(a + b) = 36
   \]
   Simplifying, we get:
   \[
   a + b = 9
   \]

4. **Balancing Colors:**
   Since the square has an equal number of black and white cells, the number of black cells covered by the figures must equal the number of white cells covered. This gives us:
   \[
   3a + b = 18 \quad \text{(black cells)}
   \]
   \[
   3b + a = 18 \quad \text{(white cells)}
   \]

5. **Solving the System of Equations:**
   From \( a + b = 9 \), we substitute \( b = 9 - a \) into the equation \( 3a + b = 18 \):
   \[
   3a + (9 - a) = 18
   \]
   Simplifying, we get:
   \[
   2a + 9 = 18
   \]
   \[
   2a = 9
   \]
   \[
   a = 4.5
   \]
   Since \( a \) must be an integer, this solution is not possible.

### Part (b): \( n = 7 \)

1. **Total Number of Cells:**
   The \( 7 \times 7 \) square has \( 49 \) cells.

2. **Divisibility by 4:**
   Each 4-cell figure covers exactly 4 cells. For the entire square to be covered by these figures, the total number of cells (49) must be divisible by 4. However:
   \[
   49 \div 4 = 12 \text{ remainder } 1
   \]
   Since 49 is not divisible by 4, it is impossible to cover the \( 7 \times 7 \) square with the given 4-cell figures.

### Conclusion:
From the analysis, we see that the only squares that can be covered by this shape are those where \( n \) is a multiple of 4.

The final answer is False for both \( \boxed{ n = 6 } \) and \( n = 7 \).