Let $m$ be a positive integer for which there exists a positive integer $n$ such that the multiplication $mn$ is a perfect square and $m- n$ is prime. Find all $m$ for $1000\leq m \leq 2021.$
1. Let \( m \) be a positive integer such that there exists a positive integer \( n \) for which \( mn \) is a perfect square and \( m - n \) is prime. We need to find all \( m \) in the range \( 1000 \leq m \leq 2021 \).

2. Since \( mn \) is a perfect square, we can write \( m \) and \( n \) in the form \( m = s q^2 \) and \( n = s r^2 \), where \( s \) is a common factor and \( q \) and \( r \) are integers. This ensures that \( mn = s^2 q^2 r^2 \) is a perfect square.

3. Given \( m - n \) is prime, we have:
   \[
   p = m - n = s q^2 - s r^2 = s (q^2 - r^2) = s (q - r)(q + r)
   \]
   Since \( p \) is prime, the factors \( s \), \( q - r \), and \( q + r \) must be such that one of them is 1. The only feasible solution is \( s = 1 \), \( q - r = 1 \), and \( q + r = p \). This simplifies to:
   \[
   q = \frac{p + 1}{2} \quad \text{and} \quad r = \frac{p - 1}{2}
   \]

4. Therefore, we have:
   \[
   m = q^2 = \left( \frac{p + 1}{2} \right)^2 \quad \text{and} \quad n = r^2 = \left( \frac{p - 1}{2} \right)^2
   \]

5. We need to find \( m \) such that \( 1000 \leq m \leq 2021 \). This translates to:
   \[
   1000 \leq \left( \frac{p + 1}{2} \right)^2 \leq 2021
   \]

6. Taking the square root of the inequality:
   \[
   \sqrt{1000} \leq \frac{p + 1}{2} \leq \sqrt{2021}
   \]
   \[
   31.62 \leq \frac{p + 1}{2} \leq 44.94
   \]

7. Multiplying through by 2:
   \[
   63.24 \leq p + 1 \leq 89.88
   \]

8. Subtracting 1 from all parts:
   \[
   62.24 \leq p \leq 88.88
   \]

9. Since \( p \) must be a prime number, we consider the prime numbers in the range \( 63 \leq p \leq 87 \):
   \[
   p \in \{67, 71, 73, 79, 83\}
   \]

10. For each \( p \), we calculate \( m \):
    - For \( p = 67 \):
      \[
      m = \left( \frac{67 + 1}{2} \right)^2 = 34^2 = 1156
      \]
    - For \( p = 71 \):
      \[
      m = \left( \frac{71 + 1}{2} \right)^2 = 36^2 = 1296
      \]
    - For \( p = 73 \):
      \[
      m = \left( \frac{73 + 1}{2} \right)^2 = 37^2 = 1369
      \]
    - For \( p = 79 \):
      \[
      m = \left( \frac{79 + 1}{2} \right)^2 = 40^2 = 1600
      \]
    - For \( p = 83 \):
      \[
      m = \left( \frac{83 + 1}{2} \right)^2 = 42^2 = 1764
      \]

11. All these values of \( m \) lie within the range \( 1000 \leq m \leq 2021 \).

The final answer is \( \boxed{\{1156, 1296, 1369, 1600, 1764\}} \)