Find all three digit numbers $n$ which are equal to the number formed by three last digit of $n^2$.
To find all three-digit numbers \( n \) such that \( n \) is equal to the number formed by the last three digits of \( n^2 \), we need to solve the congruence:
\[ n^2 \equiv n \pmod{1000} \]

1. **Rewrite the congruence:**
   \[ n^2 - n \equiv 0 \pmod{1000} \]
   \[ n(n - 1) \equiv 0 \pmod{1000} \]

2. **Factorize 1000:**
   \[ 1000 = 2^3 \cdot 5^3 \]

3. **Solve the congruence modulo 8 and 125 separately:**
   \[ n(n - 1) \equiv 0 \pmod{8} \]
   \[ n(n - 1) \equiv 0 \pmod{125} \]

4. **Solve \( n(n - 1) \equiv 0 \pmod{8} \):**
   Since \( n \) and \( n-1 \) are consecutive integers, one of them must be divisible by 2. For the product to be divisible by 8, one of them must be divisible by 8. Therefore:
   \[ n \equiv 0 \pmod{8} \quad \text{or} \quad n \equiv 1 \pmod{8} \]

5. **Solve \( n(n - 1) \equiv 0 \pmod{125} \):**
   Similarly, for the product to be divisible by 125, one of the numbers must be divisible by 125. Therefore:
   \[ n \equiv 0 \pmod{125} \quad \text{or} \quad n \equiv 1 \pmod{125} \]

6. **Combine the solutions using the Chinese Remainder Theorem:**
   We need to find \( n \) such that:
   \[ n \equiv 0 \pmod{8} \quad \text{or} \quad n \equiv 1 \pmod{8} \]
   \[ n \equiv 0 \pmod{125} \quad \text{or} \quad n \equiv 1 \pmod{125} \]

   - For \( n \equiv 0 \pmod{8} \) and \( n \equiv 0 \pmod{125} \):
     \[ n \equiv 0 \pmod{1000} \]
     This is not a three-digit number.

   - For \( n \equiv 0 \pmod{8} \) and \( n \equiv 1 \pmod{125} \):
     Let \( n = 125k + 1 \). Then:
     \[ 125k + 1 \equiv 0 \pmod{8} \]
     \[ 125k \equiv -1 \pmod{8} \]
     \[ 5k \equiv -1 \pmod{8} \]
     \[ 5k \equiv 7 \pmod{8} \]
     \[ k \equiv 7 \pmod{8} \]
     Let \( k = 8m + 7 \). Then:
     \[ n = 125(8m + 7) + 1 = 1000m + 876 \]
     So, \( n = 876 \).

   - For \( n \equiv 1 \pmod{8} \) and \( n \equiv 0 \pmod{125} \):
     Let \( n = 8k + 1 \). Then:
     \[ 8k + 1 \equiv 0 \pmod{125} \]
     \[ 8k \equiv -1 \pmod{125} \]
     \[ k \equiv -1 \cdot 8^{-1} \pmod{125} \]
     The modular inverse of 8 modulo 125 is 47 (since \( 8 \cdot 47 \equiv 1 \pmod{125} \)):
     \[ k \equiv -47 \pmod{125} \]
     \[ k \equiv 78 \pmod{125} \]
     Let \( k = 125m + 78 \). Then:
     \[ n = 8(125m + 78) + 1 = 1000m + 625 \]
     So, \( n = 625 \).

   - For \( n \equiv 1 \pmod{8} \) and \( n \equiv 1 \pmod{125} \):
     \[ n \equiv 1 \pmod{1000} \]
     This is not a three-digit number.

Therefore, the three-digit solutions are \( n = 376 \) and \( n = 625 \).

The final answer is \( \boxed{376 \text{ and } 625} \)