Solving $[x]=ax+1$ Given the equation $[x]=ax+1$, where $a \in \mathbb{R}$ and $[x]$ is the whole part of $x$, how do I solve for $x$?
 A: We solve the equation in three cases:
Case (I): If $a=1$, then $[x]=x+1$. Since $[x]\in \mathbb Z$, $x+1\in \mathbb Z$ and so $x\in \mathbb Z$. Hence, $[x]=x$ and we can conclude that $x=x+1$.
Case (II): Let $a<1$. Then, since, we have $[x]\leq x<[x]+1$, $ax+1\leq x<ax+2$. It implies that $1\leq x(1-a)<2$. Therefore, $\frac{1}{1-a}\leq x<\frac{2}{1-a}$, as $a<1$.
Case (III): Let $a<1$. Then, since, we have $[x]\leq x<[x]+1$, $ax+1\leq x<ax+2$. It implies that $1\leq x(1-a)<2$. Therefore, $\frac{1}{1-a}\geq x>\frac{2}{1-a}$, as $a>1$.
A: Let $x = h + f$, where $h\in\mathbb{Z}$ and $f\in[0,1)$. For example, for $x=-3.5$, we have $h=-4$ and $f=0.5$.
Let us assume that $-1 < x < 0$. Then $h=-1$ and $0 \lt f \lt 1$, which gives
$$
[x]=[h+f]= 0 = ah + af + 1 \\
\Rightarrow 0 = -a + af + 1 \\
\Rightarrow f = 1 - {1 \over a}.
$$
Since $0 \lt f < 1$, this implies that $a \gt 1$.
Thus, for $a \gt 1$, we can choose $h=-1$ and $f=1-{1 \over a}$, which gives
$$
\begin{array}{rcl}
[x] & = & 0 \\
ax + 1 = -a + a \left( 1-{1 \over a} \right) + 1 & = & 0,
\end{array}
$$
That is, for $a>1$, we have that $x=h+f=-1+(1-1/a)=-1/a$ is a solution.
You can similarly assume that $0 < x < 1$, which will give $h=0$ and $f=-1/a$ as solutions for $a<-1$. 
In general, we have that $h = ah + af + 1$, which implies that
$$
f = { h-1 \over a} - h
\\
\Rightarrow 0 \leq {h-1 \over a} -h < 1
\\
\Rightarrow { h-1 \over h+1} \lt a \leq {h-1 \over h} 
$$
For $h < 1$, this is either undefined or leads to a contradiction. For $h \geq 1$, however, this gives solutions. For instance, say $a=5/12$. Then we see that $h=2$ satisfies the above inequality for $a$. Solving for $f$ then gives $f=2/5$, which means that $x=h+f=2+2/5=12/5$. We can easily check that this indeed a solution:
$$
[12/5]=2 = {5 \over 12} \cdot {12 \over 5} + 1 = 2.
$$ 
Note, however, that in general there can be more than one value of $h$ that satisfies the above inequality for $a$, which means that there can be multiple solutions for those values of $a$.
You can also use Wolfram Alpha and combine the results for floor(x) = ax + 1 and ceil(x) = ax + 1 to see that there are many other solutions. Use the floor solutions when $x$ > 0 and the ceil solutions when $x<0$. 
A: Call the fractional part of $x$ with $frac(x)$ then;
$[x]=x-frac(x)=ax+1$
$\Rightarrow x-frac(x)-1=ax\Rightarrow [x-1]=ax$
$a=\frac{[x-1]}{x}$
