Let $ k\geq 2$ be  integer, $ n_1,\ n_2,\ n_3$ be positive integers and $ a_1,\ a_2,\ a_3$ be integers from $ 1,\ 2,\ \cdots ,\ k \minus{} 1$.
Let $ b_i \equal{} a_i\sum_{j \equal{} 0}^{n_i} k^{j}\ (i \equal{} 1,\ 2,\ 3)$. Find all possible pairs of integers $ (n_1,\ n_2,\ n_3)$ such that $ b_1b_2 \equal{} b_3$.
1. **Simplify the given expressions:**
   Let \( (a_1, a_2, a_3) = (a, b, c) \in \{1, 2, \ldots, k-1\} \) and \( (n_1 + 1, n_2 + 1, n_3 + 1) = (p, q, r) \) where \( p, q, r \geq 2 \). We need to find possible \( (p, q, r) \) such that:
   \[
   ab(k^p - 1)(k^q - 1) = c(k - 1)(k^r - 1)
   \]
   for some \( k \geq 2 \).

2. **Analyze the congruence modulo \( k^2 \):**
   Since \( p, q, r \geq 2 \), we have:
   \[
   ab \equiv -(k - 1)c \pmod{k^2}
   \]
   Let \( ab = -(k - 1)c + uk^2 \) for some positive integer \( u \). Then:
   \[
   u = \frac{ab + (k - 1)c}{k^2} \leq 2\left(1 - \frac{1}{k}\right)^2
   \]
   This implies \( u = 1 \), hence:
   \[
   ab = -(k - 1)c + k^2
   \]
   We also have \( ab \geq 2k - 1 > c \) since \( c \leq k - 1 \).

3. **Consider the case when \( p, q, r > 2 \):**
   Suppose all \( p, q, r > 2 \). Then:
   \[
   ab \equiv -(k - 1)c \pmod{k^3}
   \]
   By the previous equation, we have:
   \[
   -(k - 1)c + k^2 \equiv -(k - 1)c \pmod{k^3} \implies k^2 \equiv 0 \pmod{k^3}
   \]
   This is a contradiction, so at least one of \( p, q, r \) must be 2. Without loss of generality, assume \( p = 2 \) and \( r > 2 \). Then the equation becomes:
   \[
   ab(k + 1)(k^q - 1) = c(k^r - 1)
   \]

4. **Analyze the case when \( q > 2 \):**
   Suppose \( q > 2 \). Then:
   \[
   (k + 1)ab \equiv c \pmod{k^3}
   \]
   Let \( (k + 1)ab = c + vk^3 \) for some positive integer \( v \). Combining with the previous equation, we get:
   \[
   (k + 1)(-(k - 1)c + k^2) = c + vk^3
   \]
   Hence:
   \[
   v = 1 + \frac{1 - c}{k}
   \]
   So \( v = 1 \) and \( c = 1 \). Thus:
   \[
   k^2 - k + 1 = ab \leq (k - 1)^2 = k^2 - 2k + 1
   \]
   This is a contradiction, so \( q = 2 \).

5. **Solve the simplified equation:**
   The equation becomes:
   \[
   ab(k + 1)(k^2 - 1) = c(k^r - 1)
   \]
   Since \( c < ab \leq (k - 1)^2 \), we have:
   \[
   (k + 1)(k^2 - 1) = k^3 + k^2 - k - 1 < k^r - 1 \leq (k - 1)^2(k + 1)(k^2 - 1) = k^5 - k^4 - 2k^3 + 2k^2 + k - 1
   \]
   It is impossible when \( r = 3 \) or \( r \geq 5 \), so \( r = 4 \). Then the equation becomes:
   \[
   ab = c(k - 1)
   \]
   This is possible if we let \( (a, b, c) = (k - 1, 1, 1) \).

6. **Conclusion:**
   The only possible solution is \( (p, q, r) = (2, 2, 4) \), which means \( (n_1, n_2, n_3) = (1, 1, 3) \).

The final answer is \(\boxed{(1, 1, 3)}\).