Complex problem

Square the expression $S = \dfrac{1}{\sqrt{3 + 4i}} + \dfrac{1}{\sqrt{3 - 4i}}$ to get

$S^{2} = \dfrac{1}{3 + 4i} + \dfrac{1}{3 - 4i} + \dfrac{2}{\sqrt{(3 + 4i)(3 - 4i)}} =$

$\dfrac{(3 - 4i) + (3 + 4i)}{(3 + 4i)(3 - 4i)} + \dfrac{2}{\sqrt{9 + 16}} = \dfrac{6}{25} + \dfrac{2}{5} = \dfrac{16}{25}.$

and so $S = \pm \sqrt{\dfrac{16}{25}} = \pm \dfrac{4}{5}.$

Thus $a + b = 4 + 5 = \boxed{9}.$

Let $\theta = \tan^{-1} {\frac {4}{3}} \Rightarrow \cos {\theta} = \frac {3}{5}$ and $\sin {\theta} = \frac {4}{5}$ . Then, we have:

$\dfrac {1}{\sqrt{3+4i}} + \dfrac {1}{\sqrt{3-4i}} = \dfrac {1}{\sqrt{5(\frac{3}{5}+\frac {4}{5}i)}} + \dfrac {1}{\sqrt{5(\frac{3}{5}-\frac {4}{5}i)}} = \dfrac {1}{\sqrt{5e^{\theta i}}} + \dfrac {1}{\sqrt{5e^{-\theta i}}}$

$= \frac {1}{\sqrt{5}} (e^{-\frac{\theta i}{2}} + e^{\frac{\theta i}{2}}) =\frac {1}{\sqrt{5} } (\cos {\frac{\theta}{2}} - i \sin {\frac{\theta}{2}} + \cos {\frac{\theta}{2}} + i \sin {\frac{\theta}{2}} ) = \frac {2}{\sqrt{5}} \cos {\frac{\theta}{2}}$

$= \frac {2}{\sqrt{5}} \sqrt {\frac {1}{2}(\cos {\theta} + 1 )} = \frac {2}{\sqrt{5}} \sqrt {\frac {1}{2}(\frac {3}{5} + 1 )} = \frac {2}{\sqrt{5}} \sqrt {\frac {1}{2}(\frac {8}{5})} = \dfrac {4}{5}$

$\Rightarrow a+b = 4+5 =\boxed{9}$