Given an integer $n \ge 2$, determine the number of ordered $n$-tuples of integers $(a_1, a_2,...,a_n)$ such that
(a) $a_1 + a_2 + .. + a_n \ge n^2$  and
(b) $a_1^2 + a_2^2 + ... + a_n^2 \le n^3 + 1$
1. **Initial Claim and Setup**:
   We need to determine the number of ordered $n$-tuples $(a_1, a_2, \ldots, a_n)$ of integers such that:
   \[
   a_1 + a_2 + \cdots + a_n \ge n^2 \quad \text{and} \quad a_1^2 + a_2^2 + \cdots + a_n^2 \le n^3 + 1.
   \]
   We claim that $a_i = n$ for all $1 \le i \le n$ is the only solution.

2. **Applying Cauchy-Schwarz Inequality**:
   Using the Cauchy-Schwarz inequality, we have:
   \[
   \left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n 1 \right) \ge \left( \sum_{i=1}^n a_i \right)^2.
   \]
   Simplifying, we get:
   \[
   n \sum_{i=1}^n a_i^2 \ge \left( \sum_{i=1}^n a_i \right)^2.
   \]
   Given that $\sum_{i=1}^n a_i \ge n^2$, we substitute and get:
   \[
   n \sum_{i=1}^n a_i^2 \ge n^4.
   \]
   Therefore:
   \[
   \sum_{i=1}^n a_i^2 \ge n^3.
   \]

3. **Bounding $\sum_{i=1}^n a_i^2$**:
   Since $a_1^2 + a_2^2 + \cdots + a_n^2 \le n^3 + 1$, we have:
   \[
   n^3 \le \sum_{i=1}^n a_i^2 \le n^3 + 1.
   \]
   This implies:
   \[
   \sum_{i=1}^n a_i^2 \in \{n^3, n^3 + 1\}.
   \]

4. **Case Analysis**:
   - **Case 1**: $\sum_{i=1}^n a_i^2 = n^3$.
     - From the Cauchy-Schwarz inequality, equality holds if and only if $a_1 = a_2 = \cdots = a_n$. Let $a_i = k$ for all $i$. Then:
       \[
       n k^2 = n^3 \implies k^2 = n^2 \implies k = n.
       \]
       Therefore, $a_i = n$ for all $i$.

   - **Case 2**: $\sum_{i=1}^n a_i^2 = n^3 + 1$.
     - Assume there exists such an $(a_1, \ldots, a_n)$. Using the fact that $\sum_{i=1}^n a_i^2 \equiv \sum_{i=1}^n a_i \pmod{2}$, we have:
       \[
       \sum_{i=1}^n a_i \ge n^2 + 1.
       \]
       Applying Cauchy-Schwarz again:
       \[
       n(n^3 + 1) \ge (n^2 + 1)^2.
       \]
       Simplifying:
       \[
       n^4 + n \ge n^4 + 2n^2 + 1 \implies n \ge 2n^2 + 1.
       \]
       This is a contradiction for $n \ge 2$. Hence, this case is not possible.

5. **Conclusion**:
   Since the only valid case is $\sum_{i=1}^n a_i^2 = n^3$ and $a_i = n$ for all $i$, we conclude that $a_i = n$ for all $1 \le i \le n$ is the only solution.

\[
\boxed{a_i = n \text{ for all } 1 \le i \le n}
\]