Given the displacement of the particle $$x^2 = 1 + t^2$$, we want to find the acceleration, which is the second derivative of displacement with respect to time, $$a = \frac{d^2x}{dt^2}$$, and we are given that the acceleration at any time $$t$$ is $$x^{-n}$$, for us to find the value of $$n$$.

First, let's find the first derivative of displacement with respect to time, which gives us the velocity. Differentiating $$x^2 = 1 + t^2$$ with respect to t, we get:

$$2x\frac{dx}{dt} = 2t$$

This simplifies to:

$$\frac{dx}{dt} = \frac{t}{x}$$

Now, let's differentiate this velocity to find the acceleration:

$$a = \frac{d^2x}{dt^2} = \frac{d}{dt}\left(\frac{t}{x}\right)$$

To differentiate $$\frac{t}{x}$$ with respect to $$t$$, we'll use the quotient rule:

$$\frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x\cdot 1 - t\cdot \frac{dx}{dt}}{x^2}$$

Substitute $$\frac{dx}{dt} = \frac{t}{x}$$ into the equation:

$$a = \frac{x(1) - t(\frac{t}{x})}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3}$$

Recalling that the displacement equation given was $$x^2 = 1 + t^2$$, substitute this into our expression for acceleration:

$$a = \frac{1 + t^2 - t^2}{x^3} = \frac{1}{x^3}$$

Thus, the acceleration of the particle at any time $$t$$ is $$x^{-3}$$, which means our value for $$n$$ is $$3$$.