Why is $x(t)$ not periodic but $x[n]$ is? I've been stuying signals and systems and I came across this problem.
By definition, $x(t)$ denotes continuous-time signal and $x[n]$ denotes discrete-time signal.
$x(t)$ is periodic if there exists a constant $T>0$ such that $x(t) = x(t+T)$ for all $t$ is a subset of real numbers.
$x[n]$ is periodic if there exists a constant $N>0$ such that $x[n] = x[n+N]$ for all $n$ is a subset of integers.
Then I came across this question: Why is $x(t)$ aperiodic?
$x(t) = \cos((\pi t^2)/8)$
The workings I made is as follows:
$x(t+T) = \cos((\pi(t+T)^2)/8$
Assume $x(t) = x(t+T)$
i.e $(\pi  t^2)/8 + 2\pi k = (\pi(t+T)^2)/8$
$\Rightarrow t^2 + 16k = (t+T)^2
\Rightarrow 16k = T^2 + 2tT $
Considering $k$ is an integer, isn't this periodic? Please let me know if my calculation is wrong.
Apologies if I'm posting an irrelevant topic and thanks for your feedback.
 A: You have shown*:

If $x(t)$ is periodic, then there is some $T>0$ such that $\dfrac{T^2+2tT}{16}$ is an integer for every real $t$.

*Edit: As pointed out by @S.H.W in the comments, this is not quite true.  Rather, it should be

$x(t)$ is periodic if and only if there is some $T > 0$ such that at least one of $\dfrac{T^2+2tT}{16}$ or $\dfrac{T^2+2tT + 2t^2}{16}$ is an integer for every real $t.$

Since $T \neq 0$, it should be fairly apparent that there will be some $t$ such that neither of those expressions yields an integer, showing that $x(t)$ is not periodic.
To prove it, note that, for each integer $k$, there is a unique real $t$ such that $\dfrac{T^2+2tT}{16} = k$ and at most two real numbers $t$ such that $\dfrac{T^2+2tT + 2t^2}{16} = k.$  Since there are countably many integers, there are countably many $t$ such that at least one of $\dfrac{T^2+2tT}{16}$ or $\dfrac{T^2+2tT+2t^2}{16}$ is an integer.  Since there are uncountably many real numbers, there must be some real $t$ such that neither expression yields an integer.
As I mentioned above, this shows $x(t)$ is not periodic.

On the other hand, we could set e.g. $T=8$ to see that $\dfrac{T^2+2tT}{16}$ is an integer whenever $t$ is an integer, showing $x[n]$ is periodic.
A: Let $x(t) = \cos(\frac{\pi t^2}{8})$. If $x(t)$ is periodic with $T$ then there exists $T \gt 0$ such that $x(t) = x(t+T)$ for all $t \in \mathbb{R}$. So in this case we have $$\cos(\frac{\pi (t+T)^2}{8}) = \cos(\frac{\pi t^2}{8})$$If $t = 0$ then $\cos(\frac{\pi T^2}{8}) = 1$. Differentiating both sides and let $t = 0$ we have $$ T\sin(\frac{\pi T^2}{8}) = 0$$ It means $T = 0$ or $\sin(\frac{\pi T^2}{8}) = 0$. The first case isn't allowed so we conclude that $\sin(\frac{\pi T^2}{8}) = 0$. If we differentiate twice and again let $t = 0$ then $$-\frac{\pi}{16} (4 \sin(\frac{\pi T^2}{8}) + \pi T^2 \cos(\frac{\pi T^2}{8})) = 0$$
Combining the results leads to the $T = 0$ which isn't allowed according to $T \gt 0$. The motivation for using differentiation here is that $\frac{d}{dt}\cos(u(t)) = -u'(t)\sin(u(t))$ which helps us to get $T$ out of the $\cos$ function and reaches a contradiction. Of course, Brian's answer is much more elegant and it doesn't require derivative calculations.
