It's easy to see that if $0$ doesn't appear in the sequence, then all $\epsilon$ are equal. We check two cases: $I)$ If $\epsilon_k=1$ for all $k$: See that: $$a_{n+1}>a_n\iff\sqrt{a_n+1}>a_n\iff a_n\in(\overline\varphi,\varphi).$$It's easy to check that if $a_n\in(\overline\varphi,\varphi)$, then $a_{n+1}\in(\overline\varphi,\varphi)$. From this we easily conclude that if $a_n\ne\varphi$, the sequence is monotone and thus can't be periodic. $II)$ If $\epsilon_k=-1$ for all $k$: Define $b_n=-a_n$. We have $b_{n+1}=\sqrt{1-b_n}$. If $b_n\notin\{0,-\overline\varphi,1\}$ we prove that the sequence $b_{2k}$ is monotone, and thus we can't have it being periodic. First of all, see that we need to have $b_k\in(0,1)$. If $b_{n+1}>-\overline\varphi$, then: $$\sqrt{1-b_{n}}>-\overline\varphi\iff-\overline\varphi>b_n,$$and thus the sequence alternates between the intervals $(0,-\overline\varphi)$ and $(-\overline\varphi,1)$. Take some $b_n\in(0,-\overline\varphi)$. From $b_{n-1}=1-b_n^2$ and $b_{n+1}=\sqrt{1-b_n}$ we have: $$b_{n+1}>b_{n-1}\iff\sqrt{1-b_n}>1-b_n^2\iff1>(1-b_n)(1+b_n)^2\iff(b_n+\varphi)(b_n+\overline\varphi)>0,$$which is true. From this we conclude that $b_n\in\{0,-\overline\varphi,1\}$, which is easy to check that are philosophical numbers. Thus, all philosophical numbers are, $0,-1,\varphi$ and $\overline\varphi$. $\blacksquare$