[b]a)[/b] Let $ a,b $ two non-negative integers such that $ a^2>b. $ Show that the equation
$$ \left\lfloor\sqrt{x^2+2ax+b}\right\rfloor =x+a-1 $$
has an infinite number of solutions in the non-negative integers. Here, $ \lfloor\alpha\rfloor $ denotes the floor of $ \alpha. $

[b]b)[/b] Find the floor of $ m=\sqrt{2+\sqrt{2+\underbrace{\cdots}_{\text{n times}}+\sqrt{2}}} , $ where $ n $ is a natural number. Justify.
### Part (a)

1. **Given:**
   \[
   a, b \text{ are non-negative integers such that } a^2 > b.
   \]
   We need to show that the equation 
   \[
   \left\lfloor \sqrt{x^2 + 2ax + b} \right\rfloor = x + a - 1
   \]
   has an infinite number of solutions in the non-negative integers.

2. **Consider the expression inside the floor function:**
   \[
   \sqrt{x^2 + 2ax + b}
   \]
   We need to find conditions under which:
   \[
   x + a - 1 \leq \sqrt{x^2 + 2ax + b} < x + a
   \]

3. **Square both sides of the inequalities:**
   \[
   (x + a - 1)^2 \leq x^2 + 2ax + b < (x + a)^2
   \]

4. **Expand and simplify the inequalities:**
   \[
   (x + a - 1)^2 = x^2 + 2ax + a^2 - 2x - 2a + 1
   \]
   \[
   (x + a)^2 = x^2 + 2ax + a^2
   \]

5. **Set up the inequalities:**
   \[
   x^2 + 2ax + a^2 - 2x - 2a + 1 \leq x^2 + 2ax + b < x^2 + 2ax + a^2
   \]

6. **Simplify the inequalities:**
   \[
   a^2 - 2x - 2a + 1 \leq b < a^2
   \]

7. **Given \(a^2 > b\), the right inequality is always satisfied. For the left inequality:**
   \[
   b \geq a^2 - 2x - 2a + 1
   \]

8. **Rearrange to find \(x\):**
   \[
   x \leq \frac{a^2 - b - 2a + 1}{2}
   \]

9. **Since \(a^2 > b\), there are infinitely many \(x\) satisfying this inequality. Therefore, the equation has an infinite number of solutions.**

\(\blacksquare\)

### Part (b)

1. **Given:**
   \[
   m = \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}} \text{ (n times)}
   \]
   We need to find \(\lfloor m \rfloor\).

2. **Define the sequence:**
   \[
   a_n = \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}} \text{ (n times)}
   \]

3. **Base case:**
   \[
   a_1 = \sqrt{2}
   \]

4. **Inductive hypothesis:**
   Assume for \(n = k\),
   \[
   a_k = 2 \cos \left( \frac{\pi}{2^{k+1}} \right)
   \]

5. **Inductive step:**
   For \(n = k + 1\),
   \[
   a_{k+1} = \sqrt{2 + a_k}
   \]
   Using the hypothesis,
   \[
   a_{k+1} = \sqrt{2 + 2 \cos \left( \frac{\pi}{2^{k+1}} \right)}
   \]

6. **Simplify using trigonometric identity:**
   \[
   \cos(2\theta) = 2\cos^2(\theta) - 1 \implies 2\cos^2(\theta) = 1 + \cos(2\theta)
   \]
   \[
   a_{k+1} = \sqrt{2 + 2 \cos \left( \frac{\pi}{2^{k+1}} \right)} = 2 \cos \left( \frac{\pi}{2^{k+2}} \right)
   \]

7. **By induction,**
   \[
   a_n = 2 \cos \left( \frac{\pi}{2^{n+1}} \right)
   \]

8. **As \(n \to \infty\), \(\frac{\pi}{2^{n+1}} \to 0\), and \(\cos(0) = 1\):**
   \[
   a_n \to 2
   \]

9. **Therefore, for large \(n\), \(a_n\) is very close to 2, and:**
   \[
   \lfloor a_n \rfloor = 1
   \]

\(\blacksquare\)

The final answer is \( \boxed{ 1 } \)