# Periodicity of discrete time signal: $x\left [ n \right ] = \cos (\frac{\pi n^{2}}{8})$

I need to find the periodicity of the following signal:

$$x\left [ n \right ] = \cos \left(\frac{\pi n^{2}}{8}\right)$$

Now I understand that the basic procedure to determine the periodicity is to find a $$N$$ such that $$x\left [ n \right ] = x\left [ n + N\right ]$$

I applied the procedure to the aforementioned signal and got the following results:
\begin{aligned} x\left[ n\right] &= \cos \left(\frac{\pi n^{2}}{8}\right) \\ x\left[ n + N\right] &= \cos \left(\frac{\pi (n + N)^{2}}{8}\right)\\ &= \cos \left(\frac{\pi (n^{2} + 2nN + N^{2})}{8}\right)\\ &= \Re \left \{\exp \left(i \frac{\pi (n^{2} + 2nN + N^{2})}{8}\right) \right \}\\ &= \Re \left \{\exp \left(i \frac{\pi n^{2}}{8}\right)\exp \left(i \frac{\pi 2nN}{8}\right)\exp \left(i \frac{\pi N^{2}}{8}\right) \right \} \end{aligned} Now for the signal to conform to $$x\left [ n \right ] = x\left [ n + N\right ]$$: $$\Re \left \{\exp \left(i \frac{\pi 2nN}{8}\right)\exp \left(i \frac{\pi N^{2}}{8}\right) \right \} = 1$$ $$\exp(i\omega) = 1$$ only when $$\omega = 2\pi k$$ where $$k$$ is an integer. Hence:
$$\frac{\pi 2nN}{8} = \frac{\pi N^{2}}{8} = 2\pi k$$ Now logically I can see that if $$N = 8$$ the first term would be reduced to a multiple of $$2\pi$$ for all $$n$$ and hence the fundamental period would be $$N = 8$$ but how would I go about proving this mathematically, that the minimum period is indeed 8? like what set of operations would I perform on:
$$\frac{\pi 2nN}{8} = \frac{\pi N^{2}}{8} = 2\pi k$$
to obtain $$N = 8$$?

## 2 Answers

You want to prove that $\frac{nN}{8}$ is an integer for any integer $n$. Consider $n=1$. Clearly, $N$ cannot be less than 8.

You also need to prove that $\frac{N^2}{16}$ is an integer. This means that $N$ is a multiple of 4. So, the smallest $N$ that meets both conditions is 8.

You have to prove that the period $N$ is the smallest number satisfying

$$\frac{\pi}{8}(n+N)^2=\frac{\pi n^2}{8}+2\pi k,\quad k\in\mathbb{N}\tag{1}$$

From (1) you get

$$\frac{\pi}{8}n^2+\frac{\pi}{8}2nN+\frac{\pi}{8}N^2=\frac{\pi n^2}{8}+2\pi k$$

which is equivalent to

$$\frac{\pi}{8}2nN+\frac{\pi}{8}N^2=2\pi k\tag{2}$$

or

$$2nN+N^2=16k\tag{3}$$

for any value of $n$. Clearly, the smallest number $N$ for which the left-hand side of (3) is a multiple of $16$, regardless of the value of $n$, is $N=8$:

$$16n+16\cdot 4=16k$$