Find $a, b$ such that $f(x) = ax - \lfloor bx+c\rfloor$ is periodic and find its period, where $ab \ne 0$

Question

Find $a, b$ such that $f(x) = ax - \lfloor bx+c\rfloor$ is periodic and find its period, where $ab \ne 0$

I've tried to do it the following way:

$$ f(x) = f(x+T) \\ ax - \lfloor{bx + c}\rfloor = a(x+T) - \lfloor{b(x+T) + c}\rfloor \iff \\ \iff ax - \lfloor{bx + c}\rfloor - a(x+T) + \lfloor{b(x+T) + c}\rfloor = 0 \\ \lfloor{bx + c + bT}\rfloor - \lfloor{bx + c}\rfloor = aT $$

That means $aT \in \mathbb Z$, and $\lfloor{bx + c + bT}\rfloor - \lfloor{bx + c}\rfloor = aT \iff bT \in \mathbb Z$ and $bT = aT$. Therefore $a=b$ if $T\ne 0$.

But i'm stuck at finding value of $T$. Is the above correct and how do i find the period of such function?

Answer 1

The fractional part function $\{x\}=x-\lfloor x\rfloor$ is known to be periodic with period $1$.

Then

$$ax-(bx+c)+\{bx+c\}$$ can only be periodic if $a=b$ (otherwise $(a-b)x$ is aperiodic), and the period is such that $bT=1$.