$f(x)$ is a function defined on $\mathbb{R}$, with fundamental period 1. Prove that for all $f$, $g(x) = f(x^2)$ is not periodic, or give a counterexample.
If the condition "fundamental" is removed, $f(x) = \begin{cases} 1, \ x \in \mathbb{A}, \\ 0, \ \mathrm{otherwise} \end{cases}$ ($\mathbb{A}$ denotes algebraic numbers) would be a counterexample. However, it doesn't have a fundamental period.
One suggests constructing $f$ that satisfies the periodicity condition, and then prove that $f(x)$ has a fundamental period:
Choose a transcendental constant $\alpha$. Consider the equivalence relation $\sim$ on $\mathbb{R}$ generated (as a transitive closure) by $x \sim x \pm 1$ and $x^2 \sim (x \pm \alpha)^2$. Note that all equivalence classes are countable. Take $f(x) = 1$ if $x \sim 0$ and $f(x) = 0$ otherwise. Then $f$ has period $1$ and $f(x^2)$ has period $\alpha$.
But I have no idea what to do next.