#An infinite series

$\begin{aligned} S & = \sum_{x=1}^\infty \frac x \blue {x^4+4} & \small \blue{\text{By Sophie Germain identity}} \\ & = \sum_{x=1}^\infty \frac x{((x+1)^2 + 1^2)((x-1)^2+1^2)} \\ & = \sum_{x=1}^\infty \frac x{(x^2 + 2x + 2)(x^2 - 2x + 2)} \\ & = \sum_{x=1}^\infty \frac 14 \left(\frac 1{x^2 - 2x + 2} - \frac 1{x^2 + 2x + 2} \right) \\ & = \frac 14 \sum_{x=1}^\infty \left(\frac 1{(x-1)^2+1} - \frac 1{(x+1)^2 + 1} \right) \\ & = \frac 14 \left(\frac 11 - \blue{\cancel{\frac 15}} + \frac 12 - \blue{\cancel{\frac 1{10}}} + \red {\cancel{\frac 15}} - \blue{\cancel{\frac 1{17}}} + \red{\cancel{\frac 1{10}}} - \blue{\cancel{\frac 1{26}}} + \cdots \right) \\ & = \frac 14 \left(1 + \frac 12\right) = \frac 38 = \frac 6{16} \end{aligned}$

Therefore $|p-q| = |6-16| = \boxed{10}$ .

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Reference: Sophie Germain Identity