Show that the expression $(a + b + 1) (a + b - 1) (a - b + 1) (- a + b + 1)$, where $a =\sqrt{1 + x^2}$, $b =\sqrt{1 + y^2}$ and $x + y = 1$ is constant ¸and be calculated that constant value.

1. Given the expression \((a + b + 1)(a + b - 1)(a - b + 1)(-a + b + 1)\), where \(a = \sqrt{1 + x^2}\), \(b = \sqrt{1 + y^2}\), and \(x + y = 1\), we need to show that this expression is constant and calculate its value.

2. First, let's express \(a\) and \(b\) in terms of \(x\) and \(y\):
   \[
   a = \sqrt{1 + x^2}, \quad b = \sqrt{1 + y^2}
   \]

3. We are given \(x + y = 1\). Let's introduce new variables \(m\) and \(n\) such that:
   \[
   a + x = m, \quad b + y = n
   \]
   This implies:
   \[
   a = m - x, \quad b = n - y
   \]

4. Using the given \(x + y = 1\), we can write:
   \[
   a - x = \frac{1}{m}, \quad b - y = \frac{1}{n}
   \]
   This implies:
   \[
   2x = m - \frac{1}{m}, \quad 2y = n - \frac{1}{n}
   \]

5. Adding these two equations, we get:
   \[
   2(x + y) = m + n - \frac{1}{m} - \frac{1}{n}
   \]
   Since \(x + y = 1\), we have:
   \[
   2 = m + n - \frac{1}{m} - \frac{1}{n}
   \]

6. Now, let's substitute \(a\) and \(b\) back into the original expression:
   \[
   (a + b + 1)(a + b - 1)(a - b + 1)(-a + b + 1)
   \]

7. Using the substitutions \(a = m - x\) and \(b = n - y\), we get:
   \[
   (m + n)(\frac{1}{m} + \frac{1}{n})(m - \frac{1}{n})(n - \frac{1}{m})
   \]

8. Simplifying this expression:
   \[
   (m + n)\left(\frac{1}{m} + \frac{1}{n}\right)(m - \frac{1}{n})(n - \frac{1}{m})
   \]

9. We can rewrite the expression as:
   \[
   \frac{(m + n)^2 (mn - 1)^2}{(mn)^2}
   \]

10. Using the earlier result \(m + n - \frac{1}{m} - \frac{1}{n} = 2\), we have:
    \[
    (m + n - \frac{1}{m} - \frac{1}{n})^2 = 4
    \]

11. Therefore, the expression simplifies to:
    \[
    \frac{(m + n)^2 (mn - 1)^2}{(mn)^2} = 4
    \]

The final answer is \(\boxed{4}\).