We say that a real $a\geq-1$ is philosophical if there exists a sequence $\epsilon_1,\epsilon_2,\dots$, with $\epsilon_i \in\{-1,1\}$ for all $i\geq1$, such that the sequence $a_1,a_2,a_3,\dots$, with $a_1=a$, satisfies
    $$a_{n+1}=\epsilon_{n}\sqrt{a_{n}+1},\forall n\geq1$$
    and is periodic. Find all philosophical numbers.
1. **Define the problem and initial conditions:**
   We need to find all real numbers \( a \geq -1 \) that are philosophical. A number \( a \) is philosophical if there exists a sequence \( \epsilon_1, \epsilon_2, \dots \) with \( \epsilon_i \in \{-1, 1\} \) for all \( i \geq 1 \), such that the sequence \( a_1, a_2, a_3, \dots \) with \( a_1 = a \) satisfies:
   \[
   a_{n+1} = \epsilon_n \sqrt{a_n + 1}, \quad \forall n \geq 1
   \]
   and is periodic.

2. **Case 1: All \(\epsilon_k = 1\):**
   - If \(\epsilon_k = 1\) for all \( k \), then:
     \[
     a_{n+1} = \sqrt{a_n + 1}
     \]
   - We need to check if the sequence can be periodic. For the sequence to be periodic, there must exist some \( m \) such that \( a_{n+m} = a_n \) for all \( n \).
   - Consider the inequality:
     \[
     a_{n+1} > a_n \iff \sqrt{a_n + 1} > a_n \iff a_n \in (\overline{\varphi}, \varphi)
     \]
     where \(\varphi = \frac{1 + \sqrt{5}}{2}\) (the golden ratio) and \(\overline{\varphi} = \frac{1 - \sqrt{5}}{2}\) (the conjugate of the golden ratio).
   - If \( a_n \in (\overline{\varphi}, \varphi) \), then \( a_{n+1} \in (\overline{\varphi}, \varphi) \). If \( a_n \neq \varphi \), the sequence is monotone and cannot be periodic.

3. **Case 2: All \(\epsilon_k = -1\):**
   - If \(\epsilon_k = -1\) for all \( k \), then:
     \[
     a_{n+1} = -\sqrt{a_n + 1}
     \]
   - Define \( b_n = -a_n \). Then:
     \[
     b_{n+1} = \sqrt{1 - b_n}
     \]
   - We need to check if the sequence can be periodic. For the sequence to be periodic, there must exist some \( m \) such that \( b_{n+m} = b_n \) for all \( n \).
   - If \( b_n \notin \{0, -\overline{\varphi}, 1\} \), we prove that the sequence \( b_{2k} \) is monotone and thus cannot be periodic.
   - First, we need \( b_k \in (0, 1) \). 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:
     \[
     b_{n+1} > b_{n-1} \iff \sqrt{1 - b_n} > 1 - b_n^2 \iff 1 > (1 - b_n)(1 + b_n)^2 \iff (b_n + \varphi)(b_n + \overline{\varphi}) > 0
     \]
     which is true. From this, we conclude that \( b_n \in \{0, -\overline{\varphi}, 1\} \), which are easy to check as philosophical numbers.

4. **Conclusion:**
   Thus, all philosophical numbers are \( 0, -1, \varphi \), and \(\overline{\varphi}\).

The final answer is \( \boxed{ \{0, -1, \varphi, \overline{\varphi}\} } \)