We are given the expression:

$a = \sqrt{3 - \sqrt{5}} - \sqrt{3 + \sqrt{5}}$

We are asked to calculate $a^2$. Let's proceed step by step:

Step 1: Square both sides of the equation.

$a^2 = \left( \sqrt{3 - \sqrt{5}} - \sqrt{3 + \sqrt{5}} \right)^2$

Using the identity $(x - y)^2 = x^2 - 2xy + y^2$, we expand the square:

$a^2 = \left( \sqrt{3 - \sqrt{5}} \right)^2 - 2 \cdot \sqrt{3 - \sqrt{5}} \cdot \sqrt{3 + \sqrt{5}} + \left( \sqrt{3 + \sqrt{5}} \right)^2$

Step 2: Simplify the squared terms.

We can simplify the squared terms: $\left( \sqrt{3 - \sqrt{5}} \right)^2 = 3 - \sqrt{5}$ $\left( \sqrt{3 + \sqrt{5}} \right)^2 = 3 + \sqrt{5}$

So, the expression becomes: $a^2 = (3 - \sqrt{5}) + (3 + \sqrt{5}) - 2 \cdot \sqrt{(3 - \sqrt{5})(3 + \sqrt{5})}$

Step 3: Simplify the product inside the square root.

We now calculate $(3 - \sqrt{5})(3 + \sqrt{5})$, which is a difference of squares: $(3 - \sqrt{5})(3 + \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4$

Thus, we have: $a^2 = (3 - \sqrt{5}) + (3 + \sqrt{5}) - 2 \cdot \sqrt{4}$ $a^2 = 6 - 2 \cdot 2$

Step 4: Final simplification.

$a^2 = 6 - 4 = 2$

Thus, the value of $a^2$ is:

$\boxed{2}$

Would you like more details or have any questions?

Here are 5 questions related to this calculation:

1. How does the difference of squares help simplify radical expressions?
2. Can we generalize this technique for other similar expressions?
3. What are the properties of square roots that allow such simplifications?
4. How does this process relate to algebraic manipulation of irrational numbers?
5. What would happen if the inner terms under the square roots were different?

Tip: When working with radicals, always check if you can use the difference of squares to simplify expressions efficiently.