How to prove that $x(t) = \cos{(\frac{\pi}{8}\cdot t^2)}$ aperiodic?

Question

How to prove that $x(t) = \cos{(\frac{\pi}{8}\cdot t^2)}$ aperiodic?

My process was as follows:

$x(t+T)= \cos{(\frac{\pi(t+T)^2}{8})}$.

So, $T^2 + 2tT -16=0$ which seems periodic to me...

Can someone tell me how to prove it?

Answer 1

Suppose that $x$ periodic. Then its derivative wrt $t$ is also.

However $x'(t)=-\frac{1}{4} \pi x \sin(\frac{\pi}{8}x^2)$, which is obviously aperiodic because of the factor $x$.

Answer 2

Another proof by contradiction: since $\cos\frac{\pi T^2}{8}=1$, some $n\in\Bbb N$ satisfies $T=4\sqrt{n}$, whence$$-1=\cos\frac{\pi(T+\pi)^2}{8}=\cos\frac{\pi(16n+8\pi\sqrt{n}+\pi^2)}{8}\implies2n+\pi\sqrt{n}+\frac{\pi^2}{8}\in\Bbb Z\setminus2\Bbb Z.$$This contradicts the fact that $\pi$ is transcendental.