Entry 21

This infinitely nested radical is the case \(a=2\) of Entry 20 (which was the case \(a=5\)). The patterns are the same, namely \(\small{(+,+,+,-),\; (+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)}\) $$x_1=\small{+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\cdots}}}}}}}} = 2\cos\tfrac{2\pi}{17}=1.8649\dots$$ $$x_2=\small{+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\cdots}}}}}}}} = 2\cos\tfrac{4\pi}{17}=1.4780\dots$$ $$x_3=\small{+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\cdots}}}}}}}} = 2\cos\tfrac{8\pi}{17}=0.1845\dots $$ $$x_4=\small{\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}}}}} = 2\cos\tfrac{16\pi}{17}=-1.9659\dots $$ The \(x_i\) are roots of a quartic (with coefficients in \(\sqrt{17}\)). This implies that the four roots obey the system$$\begin{aligned}
x_1^2 &= x_2+2\\ x_2^2 &= x_3+2\\ x_3^2 &= x_4+2\\ x_4^2 &= x_1+2\\
\end{aligned}$$ For positive integer \(a\), apparently only \(a=2,5\) yield \(x_i\) that do not have cubic irrationalities.