Given any two positive real numbers $p\_1,p\_2$ and a function $f\_1 : \mathbb{R} \to \mathbb{R}$ such that $f\_1$ has a minimal positive period of $p\_1$ . 
Then is it true that whatever be the choice of $f\_1$ there is always a function $f\_2 : \mathbb{R} \to \mathbb{R}$ having smallest positive period of $p\_2$ such that the function $f\_1-f\_2$ is also periodic ?


A Miklos Schweitzer problem says :


Let $p\_1,p\_2$ be any two positive real numbers. Then prove that there is always functions $f\_1,f\_2 : \mathbb{R} \to \mathbb{R}$ such that $f\_i$ has minimal positive period of $p\_i$ and the function $f\_1-f\_2$ is also periodic.


In the proof the function $f\_1$ was constructed as $f\_1(x) = x$ for every real number $x \in [0,p\_1)$ and then extended periodically on $\mathbb{R}$



Given reals $p_1,p_2$ and a function $f_1$ with minimal period $p_1$.Existence a function $f_2$ with minimal period $p_2$,$f_1-f_2$ periodic?
If $p\_3$ is a period of $f\_1-f\_2$, we see that $f\_1$ must satisfy an equation $f\_1(x)+f\_1(x+p\_2+p\_3)=f\_1(x+p\_2)+f\_1(x+p\_3)$ for all $x$ (because both $f\_2$ and $f\_1-f\_2$ satisfy this equation). If, say, $f\_1(x)=\sin x$, this rewrites as $\sin(x+\frac{p\_2+p\_3}2)\cos(\frac{p\_2+p\_3}2)=\sin(x+\frac{p\_2+p\_3}2)\cos(\frac{p\_2-p\_3}2)$, $\cos(\frac{p\_2+p\_3}2)=\cos(\frac{p\_2-p\_3}2)$, $\sin\frac{p\_2}2\sin\frac{p\_3}2=0$. If $p\_2$ is not divisible by $2\pi$, we see that $p\_3$ should be divisible by $2\pi$, but in this case $f\_2=f\_1-(f\_1-f\_2)$ is $p\_3$-periodic, that in general is not so (for example, it is not so for $f\_2=\sin(x\sqrt{2})$).


