For each real number $p > 1$, find the minimum possible value of the sum $x+y$, where the numbers $x$ and $y$ satisfy the equation $(x+\sqrt{1+x^2})(y+\sqrt{1+y^2}) = p$.
1. We start with the given equation:
   \[
   (x + \sqrt{1 + x^2})(y + \sqrt{1 + y^2}) = p
   \]
   where \( p > 1 \).

2. To simplify the problem, we use the substitutions:
   \[
   x = \frac{1}{2} \left( u - \frac{1}{u} \right) \quad \text{and} \quad y = \frac{1}{2} \left( v - \frac{1}{v} \right)
   \]
   with \( u, v > 0 \).

3. Substituting these into the given equation, we get:
   \[
   \left( \frac{1}{2} \left( u - \frac{1}{u} \right) + \sqrt{1 + \left( \frac{1}{2} \left( u - \frac{1}{u} \right) \right)^2} \right) \left( \frac{1}{2} \left( v - \frac{1}{v} \right) + \sqrt{1 + \left( \frac{1}{2} \left( v - \frac{1}{v} \right) \right)^2} \right) = p
   \]

4. Simplifying the terms inside the square roots:
   \[
   \sqrt{1 + \left( \frac{1}{2} \left( u - \frac{1}{u} \right) \right)^2} = \sqrt{1 + \frac{1}{4} \left( u^2 - 2 + \frac{1}{u^2} \right)} = \sqrt{\frac{1}{4} (u^2 + 2 + \frac{1}{u^2})} = \frac{1}{2} \sqrt{u^2 + 2 + \frac{1}{u^2}}
   \]

5. Therefore, the equation becomes:
   \[
   \left( \frac{1}{2} \left( u - \frac{1}{u} \right) + \frac{1}{2} \sqrt{u^2 + 2 + \frac{1}{u^2}} \right) \left( \frac{1}{2} \left( v - \frac{1}{v} \right) + \frac{1}{2} \sqrt{v^2 + 2 + \frac{1}{v^2}} \right) = p
   \]

6. Simplifying further, we get:
   \[
   \left( \frac{1}{2} \left( u + \frac{1}{u} \right) \right) \left( \frac{1}{2} \left( v + \frac{1}{v} \right) \right) = p
   \]
   \[
   \left( u + \frac{1}{u} \right) \left( v + \frac{1}{v} \right) = 4p
   \]

7. Since \( uv = p \), we substitute \( v = \frac{p}{u} \):
   \[
   \left( u + \frac{1}{u} \right) \left( \frac{p}{u} + \frac{u}{p} \right) = 4p
   \]

8. Simplifying, we get:
   \[
   \left( u + \frac{1}{u} \right) \left( \frac{p}{u} + \frac{u}{p} \right) = 4p
   \]
   \[
   \left( u + \frac{1}{u} \right) \left( \frac{p}{u} + \frac{u}{p} \right) = 4p
   \]
   \[
   \left( u + \frac{1}{u} \right) \left( \frac{p}{u} + \frac{u}{p} \right) = 4p
   \]

9. To find the minimum value of \( x + y \), we consider:
   \[
   x + y = \frac{1}{2} \left( u - \frac{1}{u} \right) + \frac{1}{2} \left( v - \frac{1}{v} \right)
   \]
   \[
   x + y = \frac{1}{2} \left( u - \frac{1}{u} + \frac{p}{u} - \frac{u}{p} \right)
   \]

10. To minimize \( x + y \), we set \( u = \sqrt{p} \):
    \[
    x + y = \frac{1}{2} \left( \sqrt{p} - \frac{1}{\sqrt{p}} + \frac{p}{\sqrt{p}} - \frac{\sqrt{p}}{p} \right)
    \]
    \[
    x + y = \frac{1}{2} \left( \sqrt{p} - \frac{1}{\sqrt{p}} + \sqrt{p} - \frac{1}{\sqrt{p}} \right)
    \]
    \[
    x + y = \frac{1}{2} \left( 2\sqrt{p} - \frac{2}{\sqrt{p}} \right)
    \]
    \[
    x + y = \frac{1}{2} \cdot 2 \left( \sqrt{p} - \frac{1}{\sqrt{p}} \right)
    \]
    \[
    x + y = \sqrt{p} - \frac{1}{\sqrt{p}}
    \]
    \[
    x + y = \frac{p - 1}{\sqrt{p}}
    \]

The final answer is \(\boxed{\frac{p - 1}{\sqrt{p}}}\).