Let $$f : R - {0} \rightarrow (-\infty , 1)$$ be a polynomial of degree 2, satisfying $$f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$$. If $$f(K)=-2K$$, then the sum of squares of all possible values of K is:

The function $$f$$ is a polynomial of degree 2, so let $$f(x) = ax^2 + bx + c$$ with $$a \neq 0$$. The functional equation is $$f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$$.

Compute $$f\left(\frac{1}{x}\right)$$:

$$f\left(\frac{1}{x}\right) = a \left(\frac{1}{x}\right)^2 + b \left(\frac{1}{x}\right) + c = \frac{a}{x^2} + \frac{b}{x} + c$$

Substitute into the functional equation:

$$\left(ax^2 + bx + c\right) \left(\frac{a}{x^2} + \frac{b}{x} + c\right) = \left(ax^2 + bx + c\right) + \left(\frac{a}{x^2} + \frac{b}{x} + c\right)$$

Expand the left side:

$$ax^2 \cdot \frac{a}{x^2} + ax^2 \cdot \frac{b}{x} + ax^2 \cdot c + bx \cdot \frac{a}{x^2} + bx \cdot \frac{b}{x} + bx \cdot c + c \cdot \frac{a}{x^2} + c \cdot \frac{b}{x} + c \cdot c$$

$$= a^2 + abx + acx^2 + \frac{ab}{x} + b^2 + bcx + \frac{ac}{x^2} + \frac{bc}{x} + c^2$$

Group like terms:

$$acx^2 + (ab + bc)x + (a^2 + b^2 + c^2) + \frac{ab + bc}{x} + \frac{ac}{x^2}$$

The right side is:

$$ax^2 + bx + 2c + \frac{b}{x} + \frac{a}{x^2}$$

Equate coefficients for corresponding powers of $$x$$:

For $$x^2$$: $$ac = a$$   ...(1)

For $$x$$: $$ab + bc = b$$   ...(2)

For constant term: $$a^2 + b^2 + c^2 = 2c$$   ...(3)

For $$x^{-1}$$: $$ab + bc = b$$   ...(4) [same as (2)]

For $$x^{-2}$$: $$ac = a$$   ...(5) [same as (1)]

Since $$a \neq 0$$, equation (1) gives $$c = 1$$.

Substitute $$c = 1$$ into equation (2):

$$ab + b \cdot 1 = b \implies ab + b = b \implies ab = 0$$

Since $$a \neq 0$$, $$b = 0$$.

Substitute $$b = 0$$ and $$c = 1$$ into equation (3):

$$a^2 + 0^2 + 1^2 = 2 \cdot 1 \implies a^2 + 1 = 2 \implies a^2 = 1 \implies a = \pm 1$$

Thus, possible functions are:

Case 1: $$a = 1$$, $$b = 0$$, $$c = 1$$ → $$f(x) = x^2 + 1$$

Case 2: $$a = -1$$, $$b = 0$$, $$c = 1$$ → $$f(x) = 1 - x^2$$

The range of $$f$$ is given as $$(-\infty, 1)$$.

For $$f(x) = x^2 + 1$$, since $$x \neq 0$$, $$x^2 > 0$$, so $$f(x) > 1$$, which is not in $$(-\infty, 1)$$. Thus, this function is invalid.

For $$f(x) = 1 - x^2$$, since $$x \neq 0$$, $$x^2 > 0$$, so $$f(x) < 1$$. As $$x \to \infty$$, $$f(x) \to -\infty$$, so the range is $$(-\infty, 1)$$, which matches. Thus, $$f(x) = 1 - x^2$$ is valid.

Given $$f(K) = -2K$$:

$$1 - K^2 = -2K$$

Rearrange:

$$1 - K^2 + 2K = 0 \implies K^2 - 2K - 1 = 0$$

Solve the quadratic equation:

$$K = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$

So, $$K = 1 + \sqrt{2}$$ or $$K = 1 - \sqrt{2}$$.

Both values are non-zero, so they are in the domain $$\mathbb{R} - \{0\}$$.

Sum of squares:

$$(1 + \sqrt{2})^2 + (1 - \sqrt{2})^2 = (1 + 2\sqrt{2} + 2) + (1 - 2\sqrt{2} + 2) = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6$$

Thus, the sum of squares of all possible values of $$K$$ is 6.