Determine all intergers $n\geq 2$ such that $a+\sqrt{2}$ and $a^n+\sqrt{2}$ are both rational for some real number $a$ depending on $n$
1. **Base Case: \( n = 2 \)**
   - We need to find a real number \( a \) such that both \( a + \sqrt{2} \) and \( a^2 + \sqrt{2} \) are rational.
   - Let \( a = \frac{1}{2} - \sqrt{2} \).
   - Then, \( a + \sqrt{2} = \left( \frac{1}{2} - \sqrt{2} \right) + \sqrt{2} = \frac{1}{2} \), which is rational.
   - Also, \( a^2 + \sqrt{2} = \left( \frac{1}{2} - \sqrt{2} \right)^2 + \sqrt{2} = \left( \frac{1}{4} - \sqrt{2} + 2 \right) + \sqrt{2} = \frac{1}{4} + 2 = \frac{9}{4} \), which is rational.
   - Therefore, \( n = 2 \) works.

2. **General Case: \( n \geq 3 \)**
   - Assume \( n \geq 3 \) and suppose there exists a real number \( a \) such that both \( a + \sqrt{2} \) and \( a^n + \sqrt{2} \) are rational.
   - Let \( a + \sqrt{2} = l \) and \( a^n + \sqrt{2} = k \), where \( k \) and \( l \) are rational numbers.
   - Then, \( a = l - \sqrt{2} \).
   - Substituting \( a \) into \( a^n + \sqrt{2} = k \), we get:
     \[
     (l - \sqrt{2})^n + \sqrt{2} = k
     \]
   - Expanding \( (l - \sqrt{2})^n \) using the binomial theorem:
     \[
     (l - \sqrt{2})^n = \sum_{i=0}^{n} \binom{n}{i} l^{n-i} (-\sqrt{2})^i
     \]
   - Therefore:
     \[
     \sum_{i=0}^{n} \binom{n}{i} l^{n-i} (-\sqrt{2})^i + \sqrt{2} = k
     \]
   - Separating the rational and irrational parts, we get:
     \[
     \sum_{\text{even } i} \binom{n}{i} l^{n-i} (-2)^{i/2} + \left( \sum_{\text{odd } i} \binom{n}{i} l^{n-i} (-\sqrt{2})^i + \sqrt{2} \right) = k
     \]
   - For the sum to be rational, the coefficient of \( \sqrt{2} \) must be zero:
     \[
     \sum_{\text{odd } i} \binom{n}{i} l^{n-i} (-\sqrt{2})^i + \sqrt{2} = 0
     \]
   - This implies:
     \[
     \sum_{\text{odd } i} \binom{n}{i} l^{n-i} (-\sqrt{2})^i = -\sqrt{2}
     \]

3. **Case 1: \( n = 2u \) (even \( n \))**
   - For \( n = 2u \), the equation becomes:
     \[
     \sum_{i=1, \text{odd}}^{2u} \binom{2u}{i} l^{2u-i} (-\sqrt{2})^i = -\sqrt{2}
     \]
   - Simplifying, we get:
     \[
     \sum_{i=1, \text{odd}}^{2u} \binom{2u}{i} l^{2u-i} (-2)^{(i-1)/2} = 1
     \]
   - Let \( l = \frac{p}{q} \) where \( \gcd(p, q) = 1 \), \( q > 0 \), and \( p \) is an integer.
   - By the rational root theorem, \( p = 1 \) and \( q \mid 2u \).
   - However, we also get that:
     \[
     2u \cdot q^{2u-2} \cdot 2^{u-1} < q^{2u-1}
     \]
   - Combined with \( q \leq 2u \), this gives \( 2^u < 2 \), which is false for \( u > 1 \).

4. **Case 2: \( n = 2u + 1 \) (odd \( n \))**
   - For \( n = 2u + 1 \), the equation becomes:
     \[
     \sum_{i=1, \text{odd}}^{2u+1} \binom{2u+1}{i} l^{2u+1-i} (-\sqrt{2})^i = -\sqrt{2}
     \]
   - Simplifying, we get:
     \[
     \sum_{i=1, \text{odd}}^{2u+1} \binom{2u+1}{i} l^{2u+1-i} (-2)^{(i-1)/2} = 1
     \]
   - All terms being positive (since \( n \) is odd), we shall have:
     \[
     \binom{2u+1}{2u+1} \cdot 2^u < 1
     \]
   - This is vacuously false.

Hence, the only solution is \( n = 2 \), exemplified at the beginning of the solution.

The final answer is \( \boxed{ n = 2 } \)