Find all pairs $(x,y)$ of positive real numbers such that $xy$ is an integer and $x+y = \lfloor x^2 - y^2 \rfloor$.
1. **Given Equations and Initial Observations:**
   - We are given two conditions:
     \[
     xy \text{ is an integer}
     \]
     \[
     x + y = \lfloor x^2 - y^2 \rfloor
     \]
   - Since \(x\) and \(y\) are positive real numbers, \(x + y\) must be an integer because the floor function \(\lfloor x^2 - y^2 \rfloor\) is an integer.

2. **Inequalities from the Floor Function:**
   - From the floor function, we have:
     \[
     x + y \leq x^2 - y^2 < x + y + 1
     \]
   - This implies:
     \[
     x^2 - y^2 \geq x + y
     \]
     \[
     x^2 - y^2 < x + y + 1
     \]

3. **Simplifying the Inequalities:**
   - The first inequality \(x^2 - y^2 \geq x + y\) can be rewritten as:
     \[
     (x - y)(x + y) \geq x + y
     \]
     - Since \(x + y \neq 0\), we can divide both sides by \(x + y\):
     \[
     x - y \geq 1
     \]

4. **Using AM-GM Inequality:**
   - By the AM-GM inequality:
     \[
     x + y \geq 2\sqrt{xy}
     \]
   - Since \(xy\) is an integer, \(x + y \geq 2\).

5. **Combining Inequalities:**
   - From the second inequality \(x^2 - y^2 < x + y + 1\):
     \[
     (x - y)(x + y) < x + y + 1
     \]
     - Dividing both sides by \(x + y\):
     \[
     x - y < 2
     \]
   - Combining \(x - y \geq 1\) and \(x - y < 2\), we get:
     \[
     1 \leq x - y < 2
     \]

6. **Considering Possible Values for \(x - y\):**
   - Since \(x - y\) must be an integer, the only possible value is:
     \[
     x - y = 1
     \]

7. **Solving for \(x\) and \(y\):**
   - Given \(x - y = 1\) and \(x + y\) is an integer, we can solve for \(x\) and \(y\):
     \[
     x = y + 1
     \]
     - Substituting \(x = y + 1\) into \(x + y\):
     \[
     (y + 1) + y = 2y + 1
     \]
     - Since \(2y + 1\) is an integer, \(y\) must be an integer. Let \(y = d - 1\), where \(d\) is an integer:
     \[
     x = d, \quad y = d - 1
     \]
     - Therefore, the solutions are:
     \[
     (x, y) = (d, d - 1)
     \]
     - For \(d \geq 2\), these pairs satisfy the given conditions.

8. **Verification:**
   - For \(x = d\) and \(y = d - 1\):
     \[
     xy = d(d - 1) \text{ is an integer}
     \]
     \[
     x + y = d + (d - 1) = 2d - 1
     \]
     \[
     x^2 - y^2 = d^2 - (d - 1)^2 = d^2 - (d^2 - 2d + 1) = 2d - 1
     \]
     - Thus, \(x + y = \lfloor x^2 - y^2 \rfloor\) holds true.

The final answer is \( \boxed{ (x, y) = (d, d - 1) } \) where \(d\) is an integer and \(d \geq 2\).