Find all positive real numbers $c$ such that there are infinitely many pairs of positive integers $(n,m)$ satisfying the following conditions: $n \ge m+c\sqrt{m - 1}+1$ and among numbers $n. n+1,.... 2n-m$ there is no square of an integer.

(Slovakia)
To find all positive real numbers \( c \) such that there are infinitely many pairs of positive integers \( (n, m) \) satisfying the conditions \( n \ge m + c\sqrt{m - 1} + 1 \) and among the numbers \( n, n+1, \ldots, 2n - m \) there is no square of an integer, we will consider two cases: \( c \le 2 \) and \( c > 2 \).

### Case 1: \( c \le 2 \)

1. **Choose specific values for \( n \) and \( m \):**
   Let \( k \ge 1 \) be an integer. Choose \( n = k^2 + 1 \) and \( m = k^2 - 2k + 2 \).

2. **Verify the condition \( n \ge m + c\sqrt{m - 1} + 1 \):**
   \[
   m + 2\sqrt{m - 1} + 1 = (k^2 - 2k + 2) + 2(k - 1) + 1 = k^2 + 1 = n
   \]
   Since \( c \le 2 \), the inequality \( n \ge m + c\sqrt{m - 1} + 1 \) holds.

3. **Check that there are no squares among \( n, n+1, \ldots, 2n - m \):**
   \[
   2n - m = 2(k^2 + 1) - (k^2 - 2k + 2) = k^2 + 2k
   \]
   The sequence \( n, n+1, \ldots, 2n - m \) is \( k^2 + 1, k^2 + 2, \ldots, k^2 + 2k \). Since \( k^2 \) is the largest square less than \( n \) and \( (k+1)^2 = k^2 + 2k + 1 \) is greater than \( 2n - m \), there are no squares in this range.

Thus, for \( c \le 2 \), there are infinitely many pairs \( (n, m) \) satisfying the given conditions.

### Case 2: \( c > 2 \)

1. **Assume \( k^2 \) is the largest square below \( n \):**
   \[
   k^2 + 1 \le n \quad \text{and} \quad k \le \sqrt{n - 1}
   \]

2. **No squares among \( n, n+1, \ldots, 2n - m \):**
   \[
   2n - m \le k^2 + 2k
   \]

3. **Rewrite the inequality:**
   \[
   2n - m \le (\sqrt{n - 1})^2 + 2\sqrt{n - 1}
   \]
   \[
   2n - m \le n - 1 + 2\sqrt{n - 1}
   \]
   \[
   m \ge n - 2\sqrt{n - 1} - 1
   \]

4. **Compare with the given condition:**
   \[
   n \ge m + c\sqrt{m - 1} + 1
   \]
   \[
   n \le m + 2\sqrt{m - 1} + 1
   \]

Since \( c > 2 \), the inequality \( n \le m + 2\sqrt{m - 1} + 1 \) contradicts the given condition \( n \ge m + c\sqrt{m - 1} + 1 \). Therefore, there cannot be infinitely many pairs \( (n, m) \) satisfying the conditions for \( c > 2 \).

The final answer is \( \boxed{ c \le 2 } \).