Find all functions $R-->R$ such that:
$f(x^2) - f(y^2) + 2x + 1 = f(x + y)f(x - y)$
1. **Define the assertion**: Let \( P(x, y) \) be the assertion of the given functional equation:
   \[
   f(x^2) - f(y^2) + 2x + 1 = f(x + y)f(x - y)
   \]

2. **Evaluate \( P(0, 0) \)**:
   \[
   P(0, 0): f(0) - f(0) + 1 = f(0)^2 \implies 1 = f(0)^2 \implies f(0) = \pm 1
   \]

3. **Evaluate \( P\left(\frac{x}{2}, \frac{x}{2}\right) \)**:
   \[
   P\left(\frac{x}{2}, \frac{x}{2}\right): f\left(\left(\frac{x}{2}\right)^2\right) - f\left(\left(\frac{x}{2}\right)^2\right) + 2 \cdot \frac{x}{2} + 1 = f\left(\frac{x}{2} + \frac{x}{2}\right)f\left(\frac{x}{2} - \frac{x}{2}\right)
   \]
   Simplifying, we get:
   \[
   x + 1 = f(x)f(0)
   \]
   Since \( f(0) = \pm 1 \), we have two cases:
   - If \( f(0) = 1 \):
     \[
     x + 1 = f(x) \implies f(x) = x + 1
     \]
   - If \( f(0) = -1 \):
     \[
     x + 1 = -f(x) \implies f(x) = -(x + 1) = -x - 1
     \]

4. **Verify the solutions**:
   - For \( f(x) = x + 1 \):
     \[
     f(x^2) - f(y^2) + 2x + 1 = (x^2 + 1) - (y^2 + 1) + 2x + 1 = x^2 - y^2 + 2x + 1
     \]
     \[
     f(x + y)f(x - y) = (x + y + 1)(x - y + 1) = (x + y + 1)(x - y + 1) = x^2 - y^2 + x - y + x + y + 1 = x^2 - y^2 + 2x + 1
     \]
     Both sides match, so \( f(x) = x + 1 \) is a solution.

   - For \( f(x) = -x - 1 \):
     \[
     f(x^2) - f(y^2) + 2x + 1 = -(x^2 + 1) - (-(y^2 + 1)) + 2x + 1 = -x^2 - 1 + y^2 + 1 + 2x + 1 = y^2 - x^2 + 2x + 1
     \]
     \[
     f(x + y)f(x - y) = (-(x + y + 1))(-(x - y + 1)) = (x + y + 1)(x - y + 1) = x^2 - y^2 + x - y + x + y + 1 = x^2 - y^2 + 2x + 1
     \]
     Both sides match, so \( f(x) = -x - 1 \) is a solution.

Thus, the solutions to the functional equation are \( f(x) = x + 1 \) and \( f(x) = -x - 1 \).

The final answer is \( \boxed{f(x) = \pm (x + 1) \; \forall x \in \mathbb{R}} \)