A smooth or analytic function from $\mathbb{R}$ to $\mathbb{R}$ that has fixed points at $kn^2$ where $n$ is prime and $k$ is integer

Question

In A continuous function from $\mathbb{R}$ to $\mathbb{R}$ that has fixed points at $kn^2$ where $n$ is prime and $k$ is integer, I asked about continuous function. This time, I would like to ask the following variants:

1) What would a smooth function from $\mathbb{R}$ to $\mathbb{R}$ that has fixed points only at $kn^2$ for every integer $k$ and every prime number $n$ be?

2) What would an analytic function be like with all conditions mentioned in 1)? (In cases a function in 1) is not analytic)

3) What would a first-time differentiable function be like with all conditions mentioned in 1)? (if 1) is impossible to find)

It is also fine to slightly relax the condition and allow the set of fixed points to have numbers that are real numbers, but not integers.

Answer 1

Let's construct the analytic function and get it over with at once. First, what we really need is a function $f$ that vanishes precisely on the set $S$ of all non-squarefree integers. This is because having such $f$, we can form the function $x+f(x)$ which fixes the points of $S$ and nothing else.

The Weierstrass factorization theorem provides a way to construct an analytic function with prescribed zeroes $x_j$. Basically, $f(x)$ is the product of $(1-x/x_j)$ with some (non-vanishing) exponential factors inserted to improve convergence. In our case we can avoid the exponential factors by using the fact that $S$ is symmetric about $0$. Namely, let $$f(x)=x \prod_j \left(1-\frac{x^2}{x_j^2}\right)\tag1$$ where $x_j$ runs over all positive elements of $S$. Since $\sum \frac{1}{x_j^2}<\infty$, it is not hard to see that the product (1) converges uniformly on compact subsets of $\mathbb C$, and therefore defines an analytic function on $\mathbb C$ (and a fortiori on $\mathbb R$).

Note that we do need to consider the convergence on $\mathbb C$, even though we are actually interested in the behavior on $\mathbb R$ only. The limit of a locally uniformly convergent sequence of polynomials on $\mathbb R$ can be any continuous function whatsoever (by another theorem of Weierstrass). But the limit of a locally uniformly convergent sequence of polynomials on $\mathbb C$ is an analytic function. As Hadamard said,

The shortest path between two truths on the real line sometimes passes through the complex plane.