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.