Prove that $\sin(x^2+x)$ is not periodic using $f(x+T) = f(x)$

Question

Proof by contradiction:

let $\sin(x^2+x)$ be periodic, hence,

$$\sin((x+T)^2+x+T) = \sin(x^2+x)$$ $$x^2 + 2xT + T^2 + x + T- x^2 - x = 2k\pi, k \in \mathbb{Z}$$ $$2xT +T^2 + T= 2k\pi$$ $$\frac{2xT +T^2 + T}{2\pi}= k$$ which is impossible, since $x \in \mathbb{R}$,and $T$ is a constant, $2xT +T^2 + T$ can't be a multiple of $2\pi$ Is this proof correct?

Answer 1

If $f$ is periodic, then so is its derivative $f'$. If $f(x)=\sin(x+x^2)$, then $$f'(x)=(2x+1)\cos(x^2+x),$$ which is continuous and unbounded. Therefore, it's not periodic, and thus $f$ is not periodic.

Answer 2

Your basic idea is fine. However, instead of $\alpha-\beta=2\pi$ you'd have to work with something like

$\alpha-\beta$ is an even multiple of $\pi$ or $\alpha+\beta$ is an odd multiple of $\pi$

and would run into some technicalities because said multiple may even vary with $x$. It is nevertheless possible to make this work in a rigorous manner.

For example, both $\frac{\alpha-\beta}\pi=\frac{2Tx+T^2+T}\pi$ and $\frac{\alpha+\beta}{2}=\frac{2x^2+(2T+2)x+T^2+T}{\pi}$ have an integer value only for countable many values of $x$ (as long as $T\ne0$), so certainly not for all $x\in \Bbb R$.