On the Uniqueness of a Difference of Fractional Parts Let $\{ \cdot \}$ denote the fractional part function. Given $a, b, c, d \in \mathbb{R}$ such that $a \leqslant b$, $c \leqslant d$, $b - a = d - c$ and $\{ a x \} - \{ b x\} = \{c x \} - \{d x \}$ for $x \in \mathbb{Z}_{\geqslant 0}$, does it necessarily follow that there is a $\delta \in \mathbb{Z}$ such that $c = a + \delta$ and $d = b + \delta$? More generally, if $\{ a x \} - \{ b x\} = \{c x \} - \{d x \}$ for $x \in \mathbb{R}_{\geqslant 0}$, does it necessarily follow that $c = a$ and $d = b$?
I believe the answer is 'yes' in both cases, but I'm not quite sure how to prove it. If the constants are rational, then I believe a proof could follow by considering periodicity. Otherwise, if the constants are irrational, maybe a density argument would suffice.
 A: Neither of these is true as stated. The second one can easily be fixed, and I suspect the first one might be salvaged, too.
For the first one, the case where $b-a=d-c$ is an integer is a counterexample, since in this case $\{ax\}-\{bx\}=\{cx\}-\{dx\}=0$ for all $x\in\mathbb Z_{\ge0}$ regardless of whether $c-a=d-b$ is an integer. I suspect it might be true if you exclude this case, though.
For the second one, $a=b$ and $c=d$ is still a counterexample, but you can easily exclude that by requiring $a\lt b$ and $c\lt d$ instead. Then it does indeed follow that $c=a$ and $d=b$.
If a unique one of the numbers, say $d$, has highest absolute value, the jump at $x=1/|d|$ isn't cancelled, so the equation can't hold. That leaves the following cases to consider:


*

*$d=-a$: In this case $b=-c$, and the two sums $\{ax\}+\{-ax\}$ and $\{bx\}+\{-bx\}$ would cause spikes at different positions that don't cancel.

*$d=-b$: In this case either $a$ or $c$ would have greater absolute value.

*$d=-c$: In this case either $a$ or $b$ would have greater absolute value.

*$d=a$: In this case either $b$ or $c$ would have greater absolute value.


That leaves only the possibility $d=b$, and thus $c=a$.
