175. The sequence $\left(x_{n}\right)$ is defined by the conditions:

$$
x_{1}=a, \quad x_{n+1}=x_{n}^{2}
$$

Find all values of $a$ for which it is periodic.
$\triangle$ From the condition, it follows that for $n \geq 2$, $x_{n}$ is non-negative.

We will investigate the sequence for increasing and decreasing behavior. For this, we will determine the sign of the difference $x_{n+1}-x_{n}$. We will have:

$$
x_{n+1}-x_{n}=x_{n}^{2}-x_{n}=x_{n}\left(x_{n}-1\right)
$$

From this, it follows that if $x_{2}>1$, i.e., $|a|>1$, then all $x_{n}>1$. In this case, the sequence is increasing and, therefore, non-periodic.

Similarly, if $0<x_{2}<1$, i.e., $|a|<1, a \neq 0$, then the sequence is decreasing (for $-1<a<0$ starting from the second term), and thus also non-periodic.

The remaining cases are $a=0$ and $a^{2}=1$, i.e., $a= \pm 1$.

For $a=0$ and $a=1$, the sequence is constant and, therefore, periodic. For $a=-1$, we get the sequence

$$
-1 ; 1 ;-1 ; 1 ;-1 ; 1 ; \ldots
$$

which is also periodic.

Answer: $0, \pm 1$.