Let $ \left( x_n\right)_{n\ge 1} $ be a sequence having $ x_1=3 $ and defined as $ x_{n+1} =\left\lfloor \sqrt 2x_n\right\rfloor , $ for every natural number $ n. $ Find all values $ m $ for which the terms $ x_m,x_{m+1},x_{m+2} $ are in arithmetic progression, where $ \lfloor\rfloor $ denotes the integer part.
1. **Initial Sequence Analysis:**
   Given the sequence \( (x_n)_{n \ge 1} \) with \( x_1 = 3 \) and defined by \( x_{n+1} = \left\lfloor \sqrt{2} x_n \right\rfloor \), we start by calculating the first few terms:
   \[
   x_1 = 3
   \]
   \[
   x_2 = \left\lfloor \sqrt{2} \cdot 3 \right\rfloor = \left\lfloor 4.2426 \right\rfloor = 4
   \]
   \[
   x_3 = \left\lfloor \sqrt{2} \cdot 4 \right\rfloor = \left\lfloor 5.6568 \right\rfloor = 5
   \]
   \[
   x_4 = \left\lfloor \sqrt{2} \cdot 5 \right\rfloor = \left\lfloor 7.0711 \right\rfloor = 7
   \]
   \[
   x_5 = \left\lfloor \sqrt{2} \cdot 7 \right\rfloor = \left\lfloor 9.8995 \right\rfloor = 9
   \]

2. **Identifying Arithmetic Progression:**
   We need to find \( m \) such that \( x_m, x_{m+1}, x_{m+2} \) are in arithmetic progression. This means:
   \[
   x_{m+2} - x_{m+1} = x_{m+1} - x_m
   \]
   or equivalently,
   \[
   x_{m+2} = 2x_{m+1} - x_m
   \]

3. **General Formulation:**
   Let \( x_m = a \), \( x_{m+1} = n \), and \( x_{m+2} = 2n - a \). Given the sequence definition:
   \[
   x_{m+1} = \left\lfloor \sqrt{2} x_m \right\rfloor \implies n = \left\lfloor a \sqrt{2} \right\rfloor
   \]
   This implies:
   \[
   n < a \sqrt{2} < n + 1
   \]

4. **Next Term Calculation:**
   \[
   x_{m+2} = \left\lfloor \sqrt{2} x_{m+1} \right\rfloor \implies 2n - a = \left\lfloor \sqrt{2} n \right\rfloor
   \]
   This implies:
   \[
   2n - a < n \sqrt{2} < 2n - a + 1
   \]

5. **Combining Inequalities:**
   Combining the inequalities:
   \[
   2n - a < n \sqrt{2} < 2n - a + 1
   \]
   and
   \[
   n < a \sqrt{2} < n + 1
   \]
   we get:
   \[
   2n - a < n \sqrt{2} < 2a
   \]
   and thus:
   \[
   2n < 3a
   \]

6. **Further Simplification:**
   We also have:
   \[
   2n - a > n \sqrt{2} - 1 > (a \sqrt{2} - 1) \sqrt{2} - 1 = 2a - \sqrt{2} - 1
   \]
   yielding:
   \[
   2n > 3a - \sqrt{2} - 1
   \]
   Thus:
   \[
   2n \in \{3a - 1, 3a - 2\}
   \]

7. **Case Analysis:**
   For \( 2n = 3a - 1 \):
   \[
   a = 2k + 1 \implies n = 3k + 1
   \]
   Using \( n < a \sqrt{2} \):
   \[
   3k + 1 < (2k + 1) \sqrt{2}
   \]
   Simplifying:
   \[
   3k + 1 < 2k \sqrt{2} + \sqrt{2}
   \]
   \[
   k < \frac{\sqrt{2} - 1}{3 - 2\sqrt{2}}
   \]
   This yields \( a < 6 \).

8. **Verification:**
   Checking the initial terms, we see that \( x_1 = 3, x_2 = 4, x_3 = 5 \) form an arithmetic progression, and similarly, \( x_3 = 5, x_4 = 7, x_5 = 9 \) form another arithmetic progression.

The final answer is \( \boxed{ m = 1, 3 } \).