Determine all integers $m\ge2$ for which there exists an integer $n\ge1$ with 
$\gcd(m,n)=d$ and $\gcd(m,4n+1)=1$.

[i]Proposed by Gerhard Woeginger, Austria[/i]
To determine all integers \( m \ge 2 \) for which there exists an integer \( n \ge 1 \) with \(\gcd(m,n)=d\) and \(\gcd(m,4n+1)=1\), we will show that \( d \) is irrelevant and that the set \( S_d = \{ m : m \equiv 0 \pmod{d} \} \) is the answer.

1. **Initial Setup**:
   - Given \(\gcd(m,n) = d\), we can write \( m = dk \) and \( n = di \) for some integers \( k \) and \( i \).
   - We need to find \( i \) such that \(\gcd(i, k) = 1\) and \(\gcd(4di + 1, k) = 1\).

2. **Condition for \( k = 1 \)**:
   - If \( k = 1 \), then \( m = d \) and \( \gcd(m, 4n + 1) = \gcd(d, 4di + 1) = 1 \).
   - This is trivially true since \( \gcd(d, 1) = 1 \).

3. **Condition for \( k > 1 \)**:
   - Let \( p_1, p_2, \ldots, p_\ell \) be the distinct prime divisors of \( k \).
   - We need to ensure that \( \gcd(i, k) = 1 \) and \( \gcd(4di + 1, k) = 1 \).

4. **Ensuring \(\gcd(i, k) = 1\)**:
   - For each prime \( p_j \mid k \), we need \( i \not\equiv 0 \pmod{p_j} \).

5. **Ensuring \(\gcd(4di + 1, k) = 1\)**:
   - If \( p_j = 2 \), choose \( i \equiv 1 \pmod{2} \). This ensures \( i \) is odd, and \( 4di + 1 \) is odd, hence coprime with 2.
   - For \( p_j > 2 \), if \( d \) is coprime with \( p_j \), choose \( i \) such that \( 4di + 1 \equiv 2 \pmod{p_j} \). This is possible since \( p_j > 2 \).

6. **Existence of \( i \) using the Chinese Remainder Theorem**:
   - We need \( i \) such that \( i \equiv \alpha_j \pmod{p_j} \) for each \( p_j \), where \( \alpha_j \) ensures \( \gcd(4di + 1, p_j) = 1 \).
   - By the Chinese Remainder Theorem, such an \( i \) exists because the conditions are pairwise coprime.

7. **Conclusion**:
   - For any \( d \), the set \( S_d = \{ m : m \equiv 0 \pmod{d} \} \) satisfies the conditions.
   - Therefore, \( m \) must be a multiple of \( d \).

\(\blacksquare\)

The final answer is \( \boxed{ \{ m : m \equiv 0 \pmod{d} \} } \).