A four-digit number has the following properties:

(a) It is a perfect square;

(b) Its first two digits are equal

(c) Its last two digits are equal.

Find all such four-digit numbers.
1. Let the four-digit number be represented as \(aabb\), where \(a\) and \(b\) are digits.
2. Since the number is a perfect square, we can write it as \(n^2\) for some integer \(n\).
3. The number \(aabb\) can be expressed as \(1100a + 11b\), which simplifies to \(11(100a + b)\).
4. Therefore, \(n^2 = 11(100a + b)\). This implies that \(100a + b\) must be divisible by 11.
5. We can write \(100a + b = 11k\) for some integer \(k\). Thus, \(n^2 = 11 \times 11k = 121k\).
6. Since \(100a + b\) is a two-digit number, \(100a + b\) must be between 1000 and 9999. Therefore, \(100a + b\) must be between 100 and 999.
7. We need to find \(a\) and \(b\) such that \(100a + b\) is divisible by 11 and \(a\) and \(b\) are digits (0-9).

8. We also know that \(a + b \equiv 0 \pmod{11}\). Since \(a\) and \(b\) are digits, the only possible pairs \((a, b)\) that satisfy this condition are \((7, 4)\) and \((4, 7)\).

9. Let's check if these pairs form a perfect square:
   - For \((a, b) = (7, 4)\):
     \[
     100a + b = 100 \times 7 + 4 = 704
     \]
     \[
     7744 = 88^2
     \]
     Therefore, 7744 is a perfect square.

   - For \((a, b) = (4, 7)\):
     \[
     100a + b = 100 \times 4 + 7 = 407
     \]
     \[
     4774 = 69^2
     \]
     Therefore, 4774 is not a perfect square.

10. Hence, the only four-digit number that satisfies all the given conditions is 7744.

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