Determine that all $ k \in \mathbb{Z}$ such that $ \forall n$ the numbers $ 4n\plus{}1$ and $ kn\plus{}1$ have no common divisor.
1. **Understanding the problem**: We need to determine all integers \( k \) such that for all \( n \in \mathbb{Z} \), the numbers \( 4n + 1 \) and \( kn + 1 \) have no common divisor other than 1.

2. **Using the Euclidean algorithm**: We use the property of the greatest common divisor (gcd) that states:
   \[
   \gcd(a, b) = \gcd(a, a - b)
   \]
   Applying this to our problem, we need to find \( k \) such that:
   \[
   \gcd(4n + 1, kn + 1) = 1 \quad \forall n \in \mathbb{Z}
   \]

3. **Simplifying the gcd condition**: We can rewrite the gcd condition as:
   \[
   \gcd(4n + 1, kn + 1) = \gcd(4n + 1, (kn + 1) - (4n + 1)) = \gcd(4n + 1, (k - 4)n)
   \]
   Since \( n \) is arbitrary, we need:
   \[
   \gcd(4n + 1, k - 4) = 1 \quad \forall n \in \mathbb{Z}
   \]

4. **Analyzing the gcd condition**: For the gcd to be 1 for all \( n \), \( k - 4 \) must be such that it does not share any common prime factors with \( 4n + 1 \). Notice that \( 4n + 1 \) is always odd, so \( k - 4 \) must be a power of 2 to ensure that it does not share any common factors with \( 4n + 1 \).

5. **Conclusion**: Therefore, \( k - 4 \) must be a power of 2. This means:
   \[
   k = 4 + 2^m \quad \text{or} \quad k = 4 - 2^m \quad \text{for some integer } m \geq 0
   \]

The final answer is \( \boxed{ k = 4 + 2^m } \) or \( k = 4 - 2^m \) for some integer \( m \geq 0 \).