G7.1 Let $p, q, r$ be the three sides of triangle $P Q R$. If $p^{4}+q^{4}+r^{4}=2 r^{2}\left(p^{2}+q^{2}\right)$, find $a$, where $a=\cos ^{2} R$ and $R$ denotes the angle opposite $r$.
$\begin{array}{l} \cos R=\frac{p^{2}+q^{2}-r^{2}}{2 p q} \\ a=\cos ^{2} R \\ =\frac{\left(p^{2}+q^{2}-r^{2}\right)^{2}}{4 p^{2} q^{2}} \\ =\frac{p^{4}+q^{4}+r^{4}+2 p^{2} q^{2}-2 p^{2} r^{2}-2 q^{2} r^{2}}{4 p^{2} q^{2}} \\ =\frac{2 r^{2}\left(p^{2}+q^{2}\right)+2 p^{2} q^{2}-2 p^{2} r^{2}-2 q^{2} r^{2}}{4 p^{2} q^{2}} \\ =\frac{2 p^{2} q^{2}}{4 p^{2} q^{2}}=\frac{1}{2}\end{array}$