The number of actual system solutions $ \begin{cases} a^2=b+2\\ b^2=c+2 \\ c^2=a+2\\ \end{cases}$ is equal to:
Solution:
$\cos 2\theta=2\cos^2\theta-1\implies 2\cos 2\theta=(2\cos\theta)^2-2$.
Using this results in all $8$ solutions to the system.
$(2,2,2)$, $(-1,-1,-1)$, and cyclic permutations of $\left(2\cos\frac{2\pi}{7},2\cos\frac{4\pi}{7},2\cos\frac{6\pi}{7}\right)$ and $\left(2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right)$
How can i solve this in a simpler way (Without using trigonometry)
$$a^2=b+2 \tag1$$ $$b^2=c+2 \tag2$$ $$c^2=a+2 \tag3$$
By successive eliminations $$(1) \implies b=a^2-2$$ $$(2) \implies c=a^4-4a^2-2$$ $$(3) \implies a^8-8 a^6+20 a^4-16 a^2-a+2=0\tag 4$$
$(4)$ can be factorized as $$(a-2) (a+1) \left(a^3-3 a+1\right) \left(a^3+a^2-2 a-1\right)=0$$ Each cubic equation has three real roots that you can express using ... the trigonometric method !