Q:

Is $x[n]=(-1)^{n^2}$ periodic?

Is $x[n]=(-1)^{n^2}$ periodic? The answer said no, but when I draw it on a graph, it seems to be periodic, with fundamental period equal to $2$.

A:

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

