Solve $\lfloor x \rfloor = ax+1$ for integral $a$

Question

Solve $\lfloor x \rfloor = ax+1$, where $a$ is an integer.

I have found the values of $x$ for $a=0$, $a=1$ and $a=-1$. But I don't know how to continue. How can I find the solutions for integral $a\leq-2$ and $a\geq2$ ?

Answer 1

When $a=0$ there are infinite solutions , they are clearly $x\in [1,2)$. We now assume $a\neq 0$.

Since $\lfloor x \rfloor$ is an integer we conclude $ax+1$ is an integer, so $x=\frac{k}{a}$ for some $k$.

We must therefore find all the integer values for $k$ such that $\lfloor \frac{k}{a} \rfloor=k+1$.

It is clear that if $k$ is non-negative then $\lfloor \frac{k}{a} \rfloor < k+1$.

Now, if $k$ is negative we have $\lfloor \frac{k}{a} \rfloor = - \lceil \frac{-k}{a} \rceil$.

So we must find the non-negative integer values of $l$ so that $\lceil \frac{l}{a} \rceil= l-1$

$l=0$ never works

$l=1$ works if and only if $a$ is negative and $a\neq -1$.

$l=2$ works if and only if $a$ is positive and $a\neq1$.

$l=3$ works if and only if $a=2$.

$l=4$ never works.

Finally $l> 4$ does not work for any $a$, clearly not for $a=1$, and if $a\neq 1$ we have: $\lceil\frac{l}{a}\rceil\leq \frac{l}{a}+1\leq \frac{l}{2}+1< l-1$.

We conclude the solutions are as follows:

If $a<-1$ we have $x =\frac{-1}{a}$.

If $a=-1$ we have no solutions.

If $a=0$ we have $x\in(1,2]$.

If $a=1$ we have no solutions.

if $a=2$ we have $x\in \{\frac{-2}{a},\frac{-3}{a}\}$.

If $a>2$ we have $x = \frac{-2}{a}$.