Is $x[n]=(-1)^{n^2}$ periodic?
You're right, the given $x[n]$ is clearly periodic. You can show this by simply checking if

$$x[n]\stackrel{?}{=}x[n+N]\tag{1}$$

is satisfied for some positive integer $N$.

For the given $x[n]$ you get

$$\begin{align}(-1)^{n^2}&amp;\stackrel{?}{=}(-1)^{(n+N)^2}\\&amp;=(-1)^{n^2}(-1)^{N(N+2n)}\tag{2}\end{align}$$

From $(2)$ it follows that for $(1)$ (and $(2)$) to hold we require

$$(-1)^{N(N+2n)}=1\tag{3}$$

i.e., $N(N+2n)$ must be even (for arbitrary integer $n$). This is certainly the case for any even $N$, and the smallest positive even $N$ is $2$, hence the (fundamental) period of the given $x[n]$ is indeed $2$.