Suppose that for two real numbers $x$ and $y$ the following equality is true:
$$(x+ \sqrt{1+ x^2})(y+\sqrt{1+y^2})=1$$
Find (with proof) the value of $x+y$.
1. Given the equation:
   \[
   (x + \sqrt{1 + x^2})(y + \sqrt{1 + y^2}) = 1
   \]
   We will use the hyperbolic sine and cosine functions to simplify the expression. Recall the definitions:
   \[
   \sinh(a) = \frac{e^a - e^{-a}}{2}, \quad \cosh(a) = \frac{e^a + e^{-a}}{2}
   \]
   and the identity:
   \[
   \cosh^2(a) - \sinh^2(a) = 1
   \]

2. Let \( x = \sinh(a) \) and \( y = \sinh(b) \). Then:
   \[
   \sqrt{1 + x^2} = \sqrt{1 + \sinh^2(a)} = \sqrt{\cosh^2(a)} = \cosh(a)
   \]
   Similarly:
   \[
   \sqrt{1 + y^2} = \cosh(b)
   \]

3. Substitute these into the given equation:
   \[
   (x + \sqrt{1 + x^2})(y + \sqrt{1 + y^2}) = (\sinh(a) + \cosh(a))(\sinh(b) + \cosh(b))
   \]

4. Using the addition formulas for hyperbolic functions:
   \[
   \sinh(a) + \cosh(a) = e^a, \quad \sinh(b) + \cosh(b) = e^b
   \]
   Thus, the equation becomes:
   \[
   e^a \cdot e^b = e^{a+b} = 1
   \]

5. Since \( e^{a+b} = 1 \), we have:
   \[
   a + b = 0
   \]

6. Therefore:
   \[
   y = \sinh(b) = \sinh(-a) = -\sinh(a) = -x
   \]
   Hence:
   \[
   x + y = x - x = 0
   \]

The final answer is \(\boxed{0}\)