Three points $$O(0,0)$$, $$P(a, a^2)$$, $$Q(-b, b^2)$$, $$a > 0$$, $$b > 0$$, are on the parabola $$y = x^2$$. Let $$S_1$$ be the area of the region bounded by the line PQ and the parabola, and $$S_2$$ be the area of the triangle OPQ. If the minimum value of $$\frac{S_1}{S_2}$$ is $$\frac{m}{n}$$, $$\gcd(m, n) = 1$$, then $$m + n$$ is equal to:

Area $$S_2$$ (Triangle): $$\frac{1}{2} |a(b^2) - (-b)(a^2)| = \frac{1}{2}ab(a+b)$$.

Area $$S_1$$ (Parabola segment): Using the formula $$\frac{1}{6}(x_2 - x_1)^3$$ for a vertical parabola:

$$S_1 = \frac{1}{6}(a - (-b))^3 = \frac{1}{6}(a+b)^3$$.

Ratio: $$\frac{S_1}{S_2} = \frac{\frac{1}{6}(a+b)^3}{\frac{1}{2}ab(a+b)} = \frac{(a+b)^2}{3ab} = \frac{a^2 + b^2 + 2ab}{3ab} = \frac{1}{3}(\frac{a}{b} + \frac{b}{a} + 2)$$.

Minimization: By AM-GM, $$\frac{a}{b} + \frac{b}{a} \geq 2$$.

$$\min(\frac{S_1}{S_2}) = \frac{1}{3}(2 + 2) = \frac{4}{3}$$.

 $$m=4, n=3 \implies m+n = \mathbf{7}$$.