Is $f(x)=\sin(x^2)$ periodic? 
Is the function $f:\Bbb R \rightarrow \Bbb R$ defined as $f(x)=\sin(x^2)$, for all $x\in\Bbb R$, periodic?

Here's my attempt to solve this:

Let's assume that it is periodic. For a function to be periodic, it must satisfy $f(x)=f(T+x)$ for all $x\in\Bbb R$, so it must satisfy the relation for $x=0$ as well. So we get that $T^2=k\pi \iff T=\sqrt{k\pi}$, $k\in\Bbb N$ (since $T$ must be positive, we remove the $-\sqrt{k\pi}$ solution). 

So what now? I tried taking $x=\sqrt\pi$ and using the $T$ I found, and I get this: $$ \sin\pi=\sin(T+\sqrt\pi)\iff-1=\sin(\pi(\sqrt k+1)^2)\iff k+2\sqrt k+1=3/2+l  $$
Is this enough for contradiction? The left side of equation is sometimes irrational and gets rational only when $k$ is perfect square, which doesn't happen periodic, while the right hand side is always rational. Or I'm still missing some steps?
Thanks.
 A: Your approach simply forces $k$ to be a square. Observing that the LHS will be irrational sometimes is a crucial idea but it cannot apply when you are using only two intervals $[0, T]$ and $[\sqrt{\pi}, \sqrt\pi + T]$ to form equations. Here is an approach that uses more.
Assume that $f$ is periodic with period $T$. Note that the solution set for $f(x) = 0$ is $\{\pm\sqrt{n\pi}\ |\ n = 0, 1, 2, \dotsc\}$. For any nonnegative integer $m$, $f(\sqrt{m\pi}) = 0$ thus $f(\sqrt{m\pi} + T) = 0$. Hence, there exists some integer $k_m$ such that $\sqrt{m\pi} + T = \sqrt{k_m\pi}$. Note that $T = \sqrt{k_0\pi}$, which gives us
$$\begin{align}
\sqrt m + \sqrt k_0 &= \sqrt k_m\\
m + k_0 + 2\sqrt{mk_0} &= k_m.
\end{align}$$
And now we have a number-theoretic question. Let $k_0 = a^2b$, where $a^2$ is the greatest square that divides $k_0$.


*

*If $b = 1$, choose $m = 2$ to have an irrational LHS and an integer RHS.

*If $b \neq 1$, choose $m = 1$ to have an irrational LHS and an integer RHS.


Contradiction!
A: Let $f : \mathbb{R} \to \mathbb{R}$ be periodic with period $T$.


*

*The range of $f$ is precisely $f([0, T])$; in particular, if $f$ is continuous, the range of $f$ is bounded. 

*If $f$ is differentiable, then $f'$ is periodic with period $T$. 


Note that $f(x) = \sin(x^2)$ is differentiable and $f'(x) = 2x\cos(x^2)$ which is unbounded. Therefore, $f'$ cannot be periodic by the first point, and hence $f$ cannot be periodic by the second point.
