Let $$L$$ be a common tangent line to the curves $$4x^2 + 9y^2 = 36$$ and $$(2x)^2 + (2y)^2 = 31$$. Then the square of the slope of the line $$L$$ is ______.

The curve $$4x^2 + 9y^2 = 36$$ can be written as $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$, which is an ellipse with $$a^2 = 9$$ and $$b^2 = 4$$. The curve $$(2x)^2 + (2y)^2 = 31$$ simplifies to $$x^2 + y^2 = \frac{31}{4}$$, which is a circle with center at the origin and radius $$\frac{\sqrt{31}}{2}$$.

For the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$, a tangent line with slope $$m$$ has the equation $$y = mx \pm \sqrt{a^2 m^2 + b^2} = mx \pm \sqrt{9m^2 + 4}$$. The perpendicular distance from the origin to this tangent is $$\frac{\sqrt{9m^2 + 4}}{\sqrt{1 + m^2}}$$.

For the circle $$x^2 + y^2 = \frac{31}{4}$$, a tangent line with slope $$m$$ has the equation $$y = mx \pm \frac{\sqrt{31}}{2}\sqrt{1 + m^2}$$. The perpendicular distance from the origin to this tangent is $$\frac{\sqrt{31}}{2}$$.

For a common tangent, these distances must be equal (since both tangent lines pass at the same distance from the common center, the origin): $$\frac{\sqrt{9m^2 + 4}}{\sqrt{1 + m^2}} = \frac{\sqrt{31}}{2}$$.

Squaring both sides: $$\frac{9m^2 + 4}{1 + m^2} = \frac{31}{4}$$.

Cross-multiplying: $$4(9m^2 + 4) = 31(1 + m^2)$$, so $$36m^2 + 16 = 31 + 31m^2$$.

Simplifying: $$5m^2 = 15$$, hence $$m^2 = 3$$.