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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 $2$. enter image description here

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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 $2$.

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I think you're missing a critical component in your drawing/calculation

Hint: Look at the exponent

EDIT: Yep I was wrong, my bad

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    $\begingroup$ Can you explain it? It seems to me that for all odd $n$, $x[n]$ is always -1 and for all even $n$, $x[n]$ is always 1 so it is periodic. $\endgroup$ – Vinh Quang Tran Sep 15 '19 at 14:27
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Written as you did, this seems periodic to me. As you observed, the answer depends on the parity of $n$. You can rewrite $n=2\nu_n+\epsilon_n$, with $\nu_n$ integer ($\nu_n = \lfloor n/2\rfloor $), and $\epsilon_n = n-\lfloor n/2\rfloor$ is zero if $n$ is even, and $\epsilon_n$ is one if $n$ is odd. The latter series $\epsilon_n$ is $2$-periodic.

Thus, $x[n] = (-1)^{\epsilon_n}$, which is $2$-periodic as well. Sometimes, textbooks have mistakes.

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