[b]a)[/b] Let $ m,n,p\in\mathbb{Z}_{\ge 0} $ such that $ m>n $ and $ \sqrt{m} -\sqrt n=p. $ Prove that $ m $ and $ n $ are perfect squares.

[b]b)[/b] Find the numbers of four digits $ \overline{abcd} $ that satisfy the equation:
$$ \sqrt {\overline{abcd} } -\sqrt{\overline{acd}} =\overline{bb} .   $$
### Part (a)
1. Given \( m, n, p \in \mathbb{Z}_{\ge 0} \) such that \( m > n \) and \( \sqrt{m} - \sqrt{n} = p \).
2. Multiply both sides of the equation by \( \sqrt{m} + \sqrt{n} \):
   \[
   (\sqrt{m} - \sqrt{n})(\sqrt{m} + \sqrt{n}) = p(\sqrt{m} + \sqrt{n})
   \]
3. This simplifies to:
   \[
   m - n = p(\sqrt{m} + \sqrt{n})
   \]
4. Since \( p \) is an integer, \( \sqrt{m} + \sqrt{n} \) must be rational. Let \( \sqrt{m} = a \) and \( \sqrt{n} = b \) where \( a \) and \( b \) are rational numbers.
5. Therefore, \( m = a^2 \) and \( n = b^2 \). Since \( m \) and \( n \) are integers, \( a \) and \( b \) must be integers.
6. Hence, \( m \) and \( n \) are perfect squares.

\(\blacksquare\)

### Part (b)
1. Given the equation:
   \[
   \sqrt{\overline{abcd}} - \sqrt{\overline{acd}} = \overline{bb}
   \]
2. Let \( \overline{abcd} = q^2 \) and \( \overline{acd} = r^2 \). Then:
   \[
   q^2 - r^2 = 100(10a + b - a) = 100(9a + b)
   \]
3. This can be factored as:
   \[
   (q - r)(q + r) = 100(9a + b)
   \]
4. Since \( \overline{bb} = 11b \), we have:
   \[
   q - r = 11b
   \]
5. Substituting \( q - r = 11b \) into the equation:
   \[
   11b(q + r) = 100(9a + b)
   \]
6. Simplifying, we get:
   \[
   11(q + r) = \frac{100(9a + b)}{b}
   \]
7. Therefore:
   \[
   11 \mid 9a + b
   \]
8. We now check each \( b \) from 1 to 9 to find valid \( a \) and corresponding \( q \) and \( r \):

   - \( b = 1 \Rightarrow 9a + 1 \equiv 0 \pmod{11} \Rightarrow a = 6 \Rightarrow \begin{cases} q - r = 11 \\ q + r = 50 \end{cases} \Rightarrow \text{No integer solution} \)
   - \( b = 2 \Rightarrow 9a + 2 \equiv 0 \pmod{11} \Rightarrow a = 1 \Rightarrow \begin{cases} q - r = 22 \\ q + r = 50 \end{cases} \Rightarrow (q, r) = (36, 14) \)
   - \( b = 3 \Rightarrow 9a + 3 \equiv 0 \pmod{11} \Rightarrow a = 7 \Rightarrow \begin{cases} q - r = 33 \\ q + r = 200 \end{cases} \Rightarrow \text{No integer solution} \)
   - \( b = 4 \Rightarrow 9a + 4 \equiv 0 \pmod{11} \Rightarrow a = 2 \Rightarrow \begin{cases} q - r = 44 \\ q + r = 50 \end{cases} \Rightarrow \text{No integer solution} \)
   - \( b = 5 \Rightarrow 9a + 5 \equiv 0 \pmod{11} \Rightarrow a = 8 \Rightarrow \begin{cases} q - r = 55 \\ q + r = 175 \end{cases} \Rightarrow \text{No integer solution} \)
   - \( b = 6 \Rightarrow 9a + 6 \equiv 0 \pmod{11} \Rightarrow a = 3 \Rightarrow \begin{cases} q - r = 66 \\ q + r = 50 \end{cases} \Rightarrow \text{No integer solution} \)
   - \( b = 7 \Rightarrow 9a + 7 \equiv 0 \pmod{11} \Rightarrow a = 9 \Rightarrow \begin{cases} q - r = 77 \\ q + r = 50 \end{cases} \Rightarrow \text{No integer solution} \)
   - \( b = 8 \Rightarrow 9a + 8 \equiv 0 \pmod{11} \Rightarrow a = 4 \Rightarrow \begin{cases} q - r = 88 \\ q + r = 50 \end{cases} \Rightarrow \text{No integer solution} \)
   - \( b = 9 \Rightarrow 9a + 9 \equiv 0 \pmod{11} \Rightarrow a = 0 \Rightarrow \text{No integer solution} \)

9. The only valid solution is \( (q, r) = (36, 14) \) when \( b = 2 \) and \( a = 1 \).

\(\boxed{\text{Only solution is } (36, 14)}\)