Find all sets $ A$ of nonnegative integers with the property: if for the nonnegative intergers $m$ and $ n $ we have $m+n\in A$ then $m\cdot n\in A.$
To find all sets \( A \) of nonnegative integers with the property that if \( m+n \in A \) then \( m \cdot n \in A \), we will analyze the given sets and prove that they satisfy the condition, and then show that no other sets can satisfy the condition.

1. **Check the given sets:**
   - \( A = \emptyset \): This set trivially satisfies the condition because there are no elements \( m \) and \( n \) such that \( m+n \in A \).
   - \( A = \{0\} \): If \( m+n = 0 \), then \( m = 0 \) and \( n = 0 \), so \( m \cdot n = 0 \in A \).
   - \( A = \{0, 1\} \): If \( m+n \in \{0, 1\} \), then \( m \) and \( n \) can only be \( 0 \) or \( 1 \). In both cases, \( m \cdot n \in \{0, 1\} \).
   - \( A = \{0, 1, 2\} \): If \( m+n \in \{0, 1, 2\} \), then \( m \) and \( n \) can be \( 0, 1, \) or \( 2 \). The products \( m \cdot n \) are \( 0, 1, 2 \), which are all in \( A \).
   - \( A = \{0, 1, 2, 3\} \): Similarly, if \( m+n \in \{0, 1, 2, 3\} \), then \( m \) and \( n \) can be \( 0, 1, 2, \) or \( 3 \). The products \( m \cdot n \) are \( 0, 1, 2, 3 \), which are all in \( A \).
   - \( A = \{0, 1, 2, 3, 4\} \): If \( m+n \in \{0, 1, 2, 3, 4\} \), then \( m \) and \( n \) can be \( 0, 1, 2, 3, \) or \( 4 \). The products \( m \cdot n \) are \( 0, 1, 2, 3, 4 \), which are all in \( A \).
   - \( A = \mathbb{Z}_{\ge 0} \): This set trivially satisfies the condition because all nonnegative integers are included.

2. **Assume \( A \) is something else:**
   - Suppose \( n \in A \). By induction, if \( n \in A \), then \( \{0, 1, \ldots, n\} \subseteq A \). This is because if \( n \in A \), then \( (n-1) + 1 \in A \Rightarrow n-1 \in A \), and so on.
   - If \( A \) is unbounded above, then \( A = \mathbb{Z}_{\ge 0} \), which is one of the sets we have already considered.
   - Let \( \max A = a \). If \( a \le 4 \), we get one of the aforementioned sets. If \( a = 5 \) or \( a = 6 \), then \( 2 \cdot 3 = 6 \in A \) and \( 3 \cdot 3 = 9 \in A \), which is a contradiction because \( 9 > 6 \).
   - If \( a \ge 7 \), then consider \( \left\lfloor \frac{a}{2} \right\rfloor + \left\lceil \frac{a}{2} \right\rceil = a \in A \). Therefore, \( A \ni \left\lfloor \frac{a}{2} \right\rfloor \left\lceil \frac{a}{2} \right\rceil \ge \left( \frac{a}{2} - 1 \right) \left( \frac{a}{2} \right) > a \) since \( a \ge 7 \), which is a contradiction.

Thus, the only sets \( A \) that satisfy the given property are \( \emptyset \), \( \{0\} \), \( \{0, 1\} \), \( \{0, 1, 2\} \), \( \{0, 1, 2, 3\} \), \( \{0, 1, 2, 3, 4\} \), and \( \mathbb{Z}_{\ge 0} \).

The final answer is \( \boxed{ \emptyset, \{0\}, \{0, 1\}, \{0, 1, 2\}, \{0, 1, 2, 3\}, \{0, 1, 2, 3, 4\}, \mathbb{Z}_{\ge 0} } \).