Finding the maximum number of $x\in\mathbb R$ such that $a^x+b=\lfloor x\rfloor.$ 
Question : Letting $a\gt 1, b\in\mathbb R$, at most how many real number solutions does the following equation have?
$$a^x+b=\lfloor x\rfloor.$$
Here, $\lfloor x\rfloor$ is the largest integer not greater than $x$.

Motivation : I've known the following question :
Letting $c,d\in\mathbb R$, at most how many real number solutions does the following equation have?
$$x^2+cx+d=\lfloor x\rfloor.$$
We can prove that the answer is $4$. For example, $x^2-\frac{6}{5}x+\frac{3}{10}=\lfloor x\rfloor$ has $4$ real number solutions. This question got me interested in considering similar questions.
I've got that ${1.36}^x-0.4=\lfloor x\rfloor$ has $6$ real number solutions (see here). However, I can neither find any better pair $(a,b)$ nor prove that the answer is $6$. Can anyone help?
 A: $$
1.05^x + 40.45 =\lfloor x\rfloor
$$
has 12 solutions, as can be seen in the following graphic of $1.05^x + 40.45-\lfloor x\rfloor$:

My impression is that as $a\to1$, by appropriately choosing $b$ (with $b\to\infty$ as $a\to1$) the equation can have any number of solutions. I may try to prove it sometime in he future.

Theorem. For any $N\in\mathbb{N}$ there exist $a>1$ and $b>0$ such that the equation
$$
a^x+b=\lfloor x\rfloor
$$
has at least $N$ solutions.
Proof. To simplify the notation let $a=e^{\alpha}$, $\alpha>0$. Define the functions
$$
\phi(x)=e^{\alpha x}+b-\lfloor x\rfloor,\qquad \psi(x)=e^{\alpha x}+b- x.
$$
Then
$$
\phi(x)=\psi(x)+\{x\},\qquad \psi(x)\le\phi(x)\le\psi(x)+1.
$$
The function $\psi$ is strictly convex and $\lim_{x\to\pm\infty}\psi(x)=+\infty$. It has a unique minimum, attained at $x=-\ln\alpha/\alpha$, whose value is $b+(1+\ln\alpha)/\alpha$. Choose $b=-(1+\ln\alpha)/\alpha-1$. Then the minimum value of $\psi$ is -1. The equation $\psi(x)=0$ has two roots $x_1$, $x_2$ such that
$$
x_1<-\frac{\ln\alpha}{\alpha}<x_2.
$$
It is easily seen that on any interval of length $1$ contained in $[\,x_1,x_2\,]$ there is a solution of the equation $\phi(x)=0$. It follows hat there are at least $\lfloor x_2-x_1\rfloor$ solutions of $\phi(x)=0$.
It is enough to prove that $\lim_{\alpha\to0}x_2-x_1=\infty$. The change of variable $x=t-\ln\alpha/\alpha$ transforms the equation $\phi(x)=0$ into
$$
\frac{e^{\alpha t}-1}{\alpha}-t-1=0.
$$
If $t<0$ then
$$
\frac{e^{\alpha t}-1}{\alpha}< t+\frac{\alpha\,t^2}{2},
$$
which implies that
$$
x_1<-\sqrt{\frac{2}{\alpha}}.
$$
