A function with two center of symmetry - Prove that there does not exist an irrational number $q$ such that $f(ql-x)=f(ql+x)$. I would like to solve the following question.

Let $\alpha,\beta(\alpha\neq \beta)$ be real numbers.
Suppose a differentiable nonconstant function
$f:\mathbb{R}\to\mathbb{R}$ satisfies
$f(\alpha+x)=f(\alpha-x),f(\beta+x)=f(\beta-x)$ for any $x\in \mathbb{R}$. Prove the following.
(1)$f$ is a periodic function.
(2)Let $l$ be the smallest period of $f$. Prove that there does not
exist an irrational number $q$ such that $f(ql-x)=f(ql+x)$.


My attempt
(1) is solved in Functions from $\mathbb{R} \rightarrow \mathbb{R}$ with 2 centers of symmetry. (Mirror symmetry). Period is $2(\beta-\alpha)$.
I am not sure about (2). I tried to prove it with proof by contradiction. If I could find a period of the form $rl$ where $r$ is an irrational number, then by A real continuous periodic function with two incommensurate periods is constant. we have a contradiction since $f$ is nonconstant. However, I could not find such $r$.
By similar method as in (1), we can prove that $2(ql-\alpha)$ or $2(ql-\beta)$ are periods of $f$. However, I am not sure how to proceed.

Proposition $(2)$ is false.
Suppose $l=1$. If either one of $\alpha,\beta$ is irrational, then the proposition is false since you can take $q=\alpha,\beta$.
It is the case for $f(x)=\cos{\left(2\pi(x-\alpha)\right)}$ where $\alpha$ is any irrational number.
