We denote with $m_a$, $m_g$ the arithmetic mean and geometrical mean, respectively, of the positive numbers $x,y$.

a) If $m_a+m_g=y-x$, determine the value of $\dfrac xy$;

b) Prove that there exists exactly one pair of different positive integers $(x,y)$ for which $m_a+m_g=40$.
### Part (a)

1. We start with the given equation for the arithmetic mean ($m_a$) and geometric mean ($m_g$) of the positive numbers $x$ and $y$:
   \[
   m_a + m_g = y - x
   \]
   where \( m_a = \frac{x + y}{2} \) and \( m_g = \sqrt{xy} \).

2. Substitute the expressions for $m_a$ and $m_g$ into the given equation:
   \[
   \frac{x + y}{2} + \sqrt{xy} = y - x
   \]

3. Multiply the entire equation by 2 to clear the fraction:
   \[
   x + y + 2\sqrt{xy} = 2y - 2x
   \]

4. Rearrange the equation to isolate the square root term:
   \[
   x + y + 2\sqrt{xy} = 2y - 2x \implies 3x + 2\sqrt{xy} = y
   \]

5. Isolate the square root term:
   \[
   2\sqrt{xy} = y - 3x
   \]

6. Square both sides to eliminate the square root:
   \[
   4xy = (y - 3x)^2
   \]

7. Expand the right-hand side:
   \[
   4xy = y^2 - 6xy + 9x^2
   \]

8. Rearrange the equation to form a quadratic equation in $y$:
   \[
   y^2 - 10xy + 9x^2 = 0
   \]

9. Factor the quadratic equation:
   \[
   (y - 9x)(y - x) = 0
   \]

10. Solve for $y$:
    \[
    y = 9x \quad \text{or} \quad y = x
    \]

11. Since $x$ and $y$ are positive numbers and must be different, we discard the solution $y = x$. Therefore, we have:
    \[
    y = 9x
    \]

12. Thus, the ratio $\frac{x}{y}$ is:
    \[
    \frac{x}{y} = \frac{x}{9x} = \frac{1}{9}
    \]

The final answer is \(\boxed{(5, 45)}\).