The number of pairs $$a, b$$ of real numbers, such that whenever $$\alpha$$ is a root of the equation $$x^2 + ax + b = 0$$, $$\alpha^2 - 2$$ is also a root of this equation, is:

Given quadratic equation,

$$x^2+ax+b=0$$

Let its roots be

$$x_1,x_2$$

Condition given:

if $$\alpha$$ is a root, then

$$\alpha^2-2$$

is also a root of the same equation.

Thus the set of roots remains invariant under the mapping

$$f(x)=x^2-2$$

Now find all possible cases.

Case 1: Fixed points

Suppose

$$\alpha^2-2=\alpha$$

Then,

$$\alpha^2-\alpha-2=0$$

$$(\alpha-2)(\alpha+1)=0$$

Hence,

$$\alpha=2\quad \text{or}\quad \alpha=-1$$

Subcase 1(a): Roots $$\{-1,-1\}$$

Equation becomes

$$x^2+2x+1=0$$

giving

$$(a,b)=(2,1)$$

Subcase 1(b): Roots $$\{2,2\}$$

Equation becomes

$$x^2-4x+4=0$$

giving

$$(a,b)=(-4,4)$$

Subcase 1(c): Roots $$\{2,-1\}$$

Equation becomes

$$x^2-x-2=0$$

giving

$$(a,b)=(-1,-2)$$

Case 2: Two-cycle

Suppose

$$x_1^2-2=x_2$$

and

$$x_2^2-2=x_1$$

Then,

$$f(f(x))=x$$

So,

$$((x^2-2)^2-2)=x$$

$$x^4-4x^2-x+2=0$$

Factorizing,

$$x^4-4x^2-x+2=(x-2)(x+1)(x^2+x-1)$$

The roots $$2$$ and $$-1$$ are already counted.

Hence remaining roots satisfy

$$x^2+x-1=0$$

Thus equation is

$$x^2+x-1=0$$

giving

$$(a,b)=(1,-1)$$

Case 3: One root maps to a fixed point

Subcase 3(a):

Suppose one root maps to $$2.$$

Then,

$$x^2-2=2$$

$$x^2=4$$

$$x=\pm2$$

The pair $$\{2,2\}$$ is already counted, so new roots are

$$\{2,-2\}$$

Equation becomes

$$x^2-4=0$$

giving

$$(a,b)=(0,-4)$$

Subcase 3(b):

Suppose one root maps to $$-1.$$

Then,

$$x^2-2=-1$$

$$x^2=1$$

$$x=\pm1$$

The pair $$\{-1,-1\}$$ is already counted, so new roots are

$$\{-1,1\}$$

Equation becomes

$$x^2-1=0$$

giving

$$(a,b)=(0,-1)$$

Thus all possible pairs are:

$$(2,1),\ (-4,4),\ (-1,-2),\ (1,-1),\ (0,-4),\ (0,-1)$$

Therefore, total number of pairs $$(a,b)$$ is

$$\boxed{6}$$