A four digit number is called [i]stutterer[/i] if its first two digits are the same and its last two digits are also the same, e.g. $3311$ and $2222$ are stutterer numbers. Find all stutterer numbers that are square numbers.
1. Define a four-digit stutterer number as \( xxyy \), where \( x \) and \( y \) are digits from 0 to 9. This can be expressed as:
   \[
   xxyy = 1100x + 11y = 11(100x + y)
   \]
   Therefore, \( 100x + y \) must be a multiple of 11.

2. Since \( xxyy \) is also a square number, let \( n^2 = xxyy \). Then:
   \[
   n^2 = 11(100x + y)
   \]
   This implies that \( 100x + y \) must be 11 times a perfect square. Let \( k \) be an integer such that:
   \[
   100x + y = 11k^2
   \]

3. We need to find values of \( k \) such that \( 100x + y \) is a valid four-digit number. The possible values of \( k \) are those for which \( 11k^2 \) is a three-digit number (since \( 100x + y \) must be between 100 and 999). The possible values of \( k \) are:
   \[
   k = 1, 2, 3, 4, 5, 6, 7, 8, 9
   \]
   This gives us the following values for \( 11k^2 \):
   \[
   11 \cdot 1^2 = 11, \quad 11 \cdot 2^2 = 44, \quad 11 \cdot 3^2 = 99, \quad 11 \cdot 4^2 = 176, \quad 11 \cdot 5^2 = 275, \quad 11 \cdot 6^2 = 396, \quad 11 \cdot 7^2 = 539, \quad 11 \cdot 8^2 = 704, \quad 11 \cdot 9^2 = 891
   \]

4. We need \( 100x + y \) to be a valid three-digit number where \( y \) is a single digit (0-9). From the list above, the only valid value is:
   \[
   100x + y = 704
   \]
   This implies:
   \[
   x = 7 \quad \text{and} \quad y = 4
   \]

5. Therefore, the stutterer number is:
   \[
   7744
   \]

6. Verify that \( 7744 \) is a perfect square:
   \[
   7744 = 88^2
   \]

Thus, the only stutterer number that is also a square number is \( 7744 \).

The final answer is \( \boxed{ 7744 } \).