Let an integer $k > 1$ be given. For each integer $n > 1$, we put  

\[f(n) = k \cdot n \cdot \left(1-\frac{1}{p_1}\right) \cdot \left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_r}\right)\] 

where $p_1, p_2, \ldots, p_r$ are all distinct prime divisors of $n$. Find all values $k$ for which the sequence $\{x_m\}$ defined by $x_0 = a$ and $x_{m+1} = f(x_m), m = 0, 1, 2, 3, \ldots$ is bounded for all integers $a > 1$.
To solve the problem, we need to analyze the behavior of the sequence \(\{x_m\}\) defined by \(x_0 = a\) and \(x_{m+1} = f(x_m)\) for \(m = 0, 1, 2, 3, \ldots\), where \(f(n) = k \cdot n \cdot \left(1-\frac{1}{p_1}\right) \cdot \left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_r}\right)\) and \(p_1, p_2, \ldots, p_r\) are all distinct prime divisors of \(n\).

1. **Express \(f(n)\) in terms of Euler's Totient Function:**
   \[
   f(n) = k \cdot n \cdot \left(1-\frac{1}{p_1}\right) \cdot \left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_r}\right) = k \cdot \phi(n)
   \]
   where \(\phi(n)\) is Euler's Totient Function, which counts the number of integers up to \(n\) that are coprime with \(n\).

2. **Behavior of \(\phi(n)\):**
   Euler's Totient Function \(\phi(n)\) is always less than \(n\) for \(n > 1\). Specifically, \(\phi(n) = n \prod_{i=1}^r \left(1 - \frac{1}{p_i}\right)\), where \(p_1, p_2, \ldots, p_r\) are the distinct prime divisors of \(n\).

3. **Boundedness of the sequence:**
   We need to determine for which values of \(k\) the sequence \(\{x_m\}\) is bounded for all integers \(a > 1\).

4. **Case \(k = 2\):**
   \[
   f(n) = 2 \cdot \phi(n)
   \]
   For sufficiently large \(i\), all prime divisors of \(x_i\) are less than or equal to the largest prime divisor of \(k\), which is 2. Thus, \(x_i\) will eventually be a power of 2, and \(\phi(2^m) = 2^m \left(1 - \frac{1}{2}\right) = 2^{m-1}\). Therefore, \(f(2^m) = 2 \cdot 2^{m-1} = 2^m\), and the sequence becomes constant and hence bounded.

5. **Case \(k = 3\):**
   \[
   f(n) = 3 \cdot \phi(n)
   \]
   For sufficiently large \(i\), all prime divisors of \(x_i\) are less than or equal to the largest prime divisor of \(k\), which is 3. Thus, \(x_i\) will eventually be of the form \(2^{m_i}3^{l_i}\). For such \(x_i\), \(\phi(2^{m_i}3^{l_i}) = 2^{m_i-1}3^{l_i-1}(2-1)(3-1) = 2^{m_i-1}3^{l_i-1}2 = 2^{m_i}3^{l_i-1}\). Therefore, \(f(2^{m_i}3^{l_i}) = 3 \cdot 2^{m_i}3^{l_i-1} = 2^{m_i}3^{l_i}\), and the sequence becomes constant and hence bounded.

6. **Case \(k > 3\):**
   For \(k > 3\), consider the product of the terms:
   \[
   k \prod_{p \le p_0} \left(1 - \frac{1}{p}\right)
   \]
   where \(p_0\) is the largest prime divisor of \(k\). If \(k > 3\), then:
   \[
   k \prod_{p \le p_0} \left(1 - \frac{1}{p}\right) > 1
   \]
   This implies that \(f(n) = k \cdot \phi(n)\) will eventually increase \(x_i\) rather than decrease it, making the sequence unbounded.

Conclusion:
The sequence \(\{x_m\}\) is bounded for all integers \(a > 1\) if and only if \(k = 2\) or \(k = 3\).

The final answer is \( \boxed{ k = 2 } \) or \(k = 3\).