Find all real numbers $a$ s.t. $\sum_{k=0}^{2019}\left\lfloor x+\frac{k}{a}\right\rfloor=\lfloor ax\rfloor$ for every $x\in\Bbb R$.

Question

Find all real numbers $a$ wiith $a\neq 0$ and $$\left \lfloor x \right \rfloor+\left \lfloor x+\frac{1}{a} \right \rfloor+\left \lfloor x+\frac{2}{a} \right \rfloor+\ldots+\left \lfloor x+\frac{2019}{a} \right \rfloor=\left \lfloor ax \right \rfloor,$$ $\forall x\in\Bbb R$.

I know that $$\left \lfloor x \right \rfloor+\left \lfloor x+\frac{1}{a} \right \rfloor+\ldots+\left \lfloor x+\frac{a-1}{a} \right \rfloor=\left \lfloor ax \right \rfloor $$ (Hermite Identity) And from here can I make the conclusion that the only solution is 2020? I think that there might be multiple solutions. Can somebody tell me if there are other solutions? Thank you.

Answer 1

Let $$f(x)=\left [ x \right ]+\left [ x+\frac{1}{a} \right ]+\left [ x+\frac{2}{a} \right ]+...+\left [ x+\frac{2019}{a} \right ].$$

Now $$f\left(x+\frac{1}{a}\right)=\left [ x+\frac{1}{a} \right ]+\left [ x+\frac{2}{a} \right ]+...+\left [ x+\frac{2020}{a} \right ]$$

$$f\left(x+\frac{1}{a}\right)-f(x)= \left [ x+\frac{2020}{a} \right]-\left [ x \right ]$$

$$\Rightarrow \left [ a(x+\frac{1}{a})\right] -\left [ ax \right ]= \left [ x+\frac{2020}{a} \right]-\left [ x \right ]$$

$$\left [ x+\frac{2020}{a} \right]-\left [ x \right ]=1$$

Can you find the conditions now? And most importantly keep in mind it should be an identity in $x$.