For the positive integers $a, b, x_1$ we construct the sequence of numbers $(x_n)_{n=1}^{\infty}$ such that  $x_n = ax_{n-1} + b$ for each $n \ge  2$. Specify the conditions for the given numbers $a, b$ and $x_1$ which are necessary and sufficient for all indexes $m, n$ to apply the implication $m | n \Rightarrow x_m | x_n$.

(Jaromír Šimša)
1. **Initial Setup and Sequence Definition:**
   We are given a sequence \((x_n)_{n=1}^{\infty}\) defined by the recurrence relation:
   \[
   x_n = ax_{n-1} + b \quad \text{for each } n \ge 2
   \]
   with initial term \(x_1\). We need to find conditions on \(a\), \(b\), and \(x_1\) such that for all indices \(m\) and \(n\), the implication \(m \mid n \Rightarrow x_m \mid x_n\) holds.

2. **Assumption and Induction:**
   Assume \(x_1 = 1\) (without loss of generality, as we can scale the sequence by \(x_1\)). We aim to prove by induction that:
   \[
   x_n = a^{n-1} + b \frac{a^{n-1} - 1}{a - 1}
   \]
   - **Base Case:** For \(n = 1\), \(x_1 = 1\).
   - **Inductive Step:** Assume the formula holds for \(n = k\). Then for \(n = k+1\):
     \[
     x_{k+1} = ax_k + b = a \left(a^{k-1} + b \frac{a^{k-1} - 1}{a - 1}\right) + b = a^k + b \frac{a^k - a + a - 1}{a - 1} = a^k + b \frac{a^k - 1}{a - 1}
     \]
     Thus, the formula holds for all \(n\) by induction.

3. **Divisibility Condition:**
   Fix \(n\) and take \(m\) such that \(m \mid n\). We need:
   \[
   x_m \mid x_n \quad \text{or} \quad a^{m-1} + b \frac{a^{m-1} - 1}{a - 1} \mid a^{n-1} + b \frac{a^{n-1} - 1}{a - 1}
   \]
   This implies:
   \[
   a^{m-1} + b \frac{a^{m-1} - 1}{a - 1} \mid b \frac{a^{n-1} - a^{m-1}}{a - 1}
   \]

4. **Special Case \(m = 2n\):**
   Choosing \(m = 2n\) and noting that \(\frac{a^n - 1}{a - 1} \mid \frac{a^{kn} - 1}{a - 1}\) for any \(k\), we get:
   \[
   a^{n-1} + b \frac{a^{n-1} - 1}{a - 1} \mid b \frac{a^n - 1}{a - 1}
   \]
   Subtracting \(a\) times the left side from the right side, we need:
   \[
   a^{n-1} + b \frac{a^{n-1} - 1}{a - 1} \mid a^n - b
   \]

5. **General Condition:**
   For the above to hold for all \(n\), we must have:
   \[
   a^{n-1} + b \frac{a^{n-1} - 1}{a - 1} \mid a^n - b
   \]
   This implies:
   \[
   a^{n-1}(a^2 - (k+1)a - kb) = ba - (b + k - kb)
   \]
   For this to hold for infinitely many \(n\), the left-hand side must grow arbitrarily large unless:
   \[
   a^2 - (k+1)a - kb = 0
   \]
   Solving this quadratic equation, we find:
   \[
   a = b = 1
   \]

6. **Conclusion:**
   If \(a = b = 1\), then \(x_n = n\), which satisfies the condition \(m \mid n \Rightarrow x_m \mid x_n\). Undoing the assumption \(x_1 = 1\), we get that \((a, b, x_1) = (1, t, t)\) for any positive integer \(t\).

The final answer is \( \boxed{ (a, b, x_1) = (1, t, t) } \) for any positive integer \(t\).