\( \begin{array}{c}\left(a^{4}+\frac{1}{a^{4}}\right)=(\sqrt{2+\sqrt{2}})^{2} \\ a^{8}+2\left(a^{4}\right)\left(\frac{1}{a^{4}}\right)+\frac{1}{a^{8}}=2+\sqrt{2} \\ a^{8}+\frac{1}{a^{8}}+2=2+\sqrt{2} \\ a^{8}+\frac{1}{a^{8}}=\sqrt{2} \\ \left(a^{8}+\frac{1}{a^{8}}\right)^{2}=2 \Rightarrow a^{16}+2+\frac{1}{a^{46}}=2 \\ \Rightarrow a^{16}+\frac{1}{a_{16}}=0\end{array} \)
Answer
\( \begin{array}{c} \left(a^{8}+\frac{1}{a^{8}}\right)^{2}=2 \Rightarrow a^{16}+2+\frac{1}{a^{16}}=2 \end{array} \)
Step 1 ：\( \begin{array}{c} a^{8}+\frac{1}{a^{8}}+2=2+\sqrt{2} \end{array} \)
Step 2 ：\( \begin{array}{c} a^{8}+\frac{1}{a^{8}}=\sqrt{2} \end{array} \)
Step 3 ：\( \begin{array}{c} \left(a^{8}+\frac{1}{a^{8}}\right)^{2}=2 \Rightarrow a^{16}+2+\frac{1}{a^{16}}=2 \end{array} \)
