Question 2 Let $r_{1}, r_{2}, \cdots, r_{20}$ be the roots of the polynomial $x^{20}-7 x^{3}+1$. If $\frac{1}{r_{1}^{2}+1}+\frac{1}{r_{2}^{2}+1}+\cdots+\frac{1}{r_{20}^{2}+1}$ can be expressed in the form $\frac{m}{n}((m, n)=1)$, find the value of $m+n$. [4]
(2016, Carnegie Mellon University Informatics and Mathematics Competition)
$$
\begin{array}{l}
\text { Given } \frac{1}{r_{k}^{2}+1}=\frac{1}{\left(r_{k}+\mathrm{i}\right)\left(r_{k}-\mathrm{i}\right)} \\
=\frac{1}{2 \mathrm{i}}\left(\frac{1}{r_{k}-\mathrm{i}}-\frac{1}{r_{k}+\mathrm{i}}\right) \\
=\frac{1}{2 \mathrm{i}}\left(\frac{1}{-\left(\mathrm{i}-r_{k}\right)}+\frac{1}{-\mathrm{i}-r_{k}}\right) \\
=\frac{1}{2 \mathrm{i}}\left(\frac{1}{-\mathrm{i}-r_{k}}-\frac{1}{\mathrm{i}-r_{k}}\right) \text {, } \\
\text { then } \frac{1}{r_{1}^{2}+1}+\frac{1}{r_{2}^{2}+1}+\cdots+\frac{1}{r_{20}^{2}+1} \\
=\frac{1}{2 \mathrm{i}}\left(\sum_{k=1}^{20} \frac{1}{-\mathrm{i}-r_{k}}-\sum_{k=1}^{20} \frac{1}{\mathrm{i}-r_{k}}\right) \\
=\frac{1}{2 \mathrm{i}}\left(\frac{P^{\prime}(-\mathrm{i})}{P(-\mathrm{i})}-\frac{P^{\prime}(\mathrm{i})}{P(\mathrm{i})}\right) \text {. } \\
\text { Let } P(x)=x^{20}-7 x^{3}+1 \text {. Hence } \\
P(\mathrm{i})=\mathrm{i}^{20}-7 \mathrm{i}^{3}+1=2+7 \mathrm{i} \text {, } \\
P(-\mathrm{i})=(-\mathrm{i})^{20}-7(-\mathrm{i})^{3}+1=2-7 \mathrm{i} \text {, } \\
P^{\prime}(x)=20 x^{19}-21 x^{2} \text {. } \\
\text { Thus } P^{\prime}(i)=20 i^{19}-21 i^{2}=21-20 i \text {, } \\
P^{\prime}(-\mathrm{i})=20(-\mathrm{i})^{19}-21(-\mathrm{i})^{2}=21+20 \mathrm{i} \text {. } \\
\text { Therefore, } \frac{1}{r_{1}^{2}+1}+\frac{1}{r_{2}^{2}+1}+\cdots+\frac{1}{r_{20}^{2}+1} \\
=\frac{1}{2 \mathrm{i}}\left(\frac{P^{\prime}(-\mathrm{i})}{P(-\mathrm{i})}-\frac{P^{\prime}(\mathrm{i})}{P(\mathrm{i})}\right) \\
=\frac{1}{2 \mathrm{i}}\left(\frac{21+20 \mathrm{i}}{2+7 \mathrm{i}}-\frac{21-20 \mathrm{i}}{2-7 \mathrm{i}}\right) \\
=\frac{187}{53} \text {. } \\
\end{array}
$$

Therefore, $m+n=187+53=240$.