 Determine all positive integers $n$ such that $4k^2 +n$ is a prime number for all non-negative integer $k$ smaller than $n$.
1. **Initial Observation**: We need to determine all positive integers \( n \) such that \( 4k^2 + n \) is a prime number for all non-negative integers \( k \) smaller than \( n \). 

2. **Prime Condition for \( n \)**: 
   - Taking \( k = 0 \), we get \( 4 \cdot 0^2 + n = n \). 
   - Therefore, \( n \) itself must be a prime number.

3. **Oddness of \( n \)**:
   - Since \( n \) is a prime number greater than 2, \( n \) must be odd.

4. **Mersenne Prime Hypothesis**:
   - We hypothesize that \( n \) is a Mersenne prime, i.e., \( n = 2^p - 1 \) for some prime \( p \).

5. **Divisibility Argument**:
   - Let \( r \) be an odd prime divisor of \( n + 1 \). Then \( r \leq \frac{n+1}{2} < n \).
   - There exists a \( k \) such that \( r \mid 4k^2 - 1 \). This can be achieved by choosing \( k \equiv \frac{1}{2} \pmod{r} \).

6. **Contradiction for \( k \)**:
   - For this \( k \), \( r \mid 4k^2 + n \). Since \( 4k^2 + n > r \), \( 4k^2 + n \) cannot be a prime, leading to a contradiction.
   - Hence, \( n \) must be a Mersenne prime.

7. **Manual Check for Small \( p \)**:
   - For \( p = 2 \), \( n = 2^2 - 1 = 3 \).
   - For \( p = 3 \), \( n = 2^3 - 1 = 7 \).

8. **Exclusion of Larger \( p \)**:
   - Assume \( p \geq 5 \). Then \( n = 2^p - 1 \).
   - Consider \( n + 9 = 2^p + 8 = 8(2^{p-3} + 1) \).
   - Since \( 2^{p-3} + 1 \) is odd, there exists a prime \( r \mid 2^{p-3} + 1 \).
   - \( r \mid \frac{n + 9}{8} < n \), so \( r \leq n - 1 \).
   - There exists a \( k \) such that \( 4k^2 \equiv 9 \pmod{r} \). This can be achieved by choosing \( k \equiv \frac{3}{2} \pmod{r} \).
   - For this \( k \), \( r \mid 4k^2 + n \) but \( 4k^2 + n > r \), leading to a contradiction.

9. **Conclusion**:
   - Therefore, the only possible values for \( n \) are the Mersenne primes for \( p < 5 \), which are \( n = 3 \) and \( n = 7 \).

The final answer is \( \boxed{ n = 3, 7 } \).