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?
$$ 1.05^x + 40.45 =\lfloor x\rfloor $$ 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 $$ x_1<-\sqrt{\frac{2}{\alpha}}. $$