Let the parabola y = x^2 + px + q passing through the point (1, -1)...

Given the equation of the parabola $y = x^2 + px + q$.

Since it passes through $(1, -1)$, we have:

$-1 = 1 + p + q \Rightarrow q = -p - 2$

The vertex of the parabola $y = x^2 + px + q$ occurs at $x = -\dfrac{p}{2}$.

The $y$-coordinate of the vertex is:

$y_v = \left(-\dfrac{p}{2}\right)^2 + p\left(-\dfrac{p}{2}\right) + q = -\dfrac{p^2}{4} + q$

The distance between the vertex and the $x$-axis is $|y_v|$. Substituting $q = -p - 2$ into the expression for $y_v$:

$|y_v| = \left|-\dfrac{p^2}{4} - p - 2\right| = \dfrac{1}{4}|p^2 + 4p + 8| = \dfrac{1}{4}|(p+2)^2 + 4|$

To minimize this distance, the term $(p+2)^2$ must be minimum. Since $(p+2)^2 \ge 0$ for all real $p$, the minimum value occurs at $p = -2$.

Substituting $p = -2$ into $q = -p - 2$, we get:

$q = -(-2) - 2 = 0$

Therefore, the value of $p^2 + q^2$ is:

$p^2 + q^2 = (-2)^2 + 0^2 = 4$

Answer: $4$