Properties of $f$ such that $f(x-1)-f(x+1)=\lfloor f(x)-f(x+1) \rfloor$

Question

Let $f\colon\mathbb{R}\to\mathbb{R}$ be such that $f(x-1)-f(x+1)=\lfloor f(x)-f(x+1) \rfloor$ for all $x$. Then $f$ must be:

a) periodic

b) constant

c) discontinuous if nonconstant

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I do not have an example of such a non-constant function.

My attempt: using $t-1<\lfloor t\rfloor\le t$ I obtained $f(x)-f(x+1)-1 < f(x-1)-f(x+1) \le f(x)-f(x+1)$ -- equivalent to $0\le f(x)-f(x-1)<1$. This is for all $x$, so $x:=x+1$ gives $0\le f(x+1)-f(x)<1$ or $-1<f(x)-f(x+1)\le 0$. Hence $\lfloor f(x)-f(x+1) \rfloor$ is always $0$ or $-1$, so I have $f(x-1)-f(x+1)\in\{-1,0\}$ always, but now what?

Answer 1

HINT: Show that any $f:\Bbb R\to\Bbb R$ that is periodic with period $1$ has the desired property; that will give you continuous, non-constant examples.

Answer 2

$f(x) = x/2$ is an example that is continuous and neither periodic nor constant.