Representing positive integers as floor of integer powers of real number.

Question

Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $\lfloor{a^b}\rfloor = c$?

Answer 1

No.

The simplest way to see why the answer is no is the fact that, for $a>1$, you have

$$\lim_{b\to\infty} (a^b-a^{b-1}) = \infty$$

(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)

The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $\lfloor a^{b-1}\rfloor$ and $\lfloor a^b\rfloor$ (that is, $\lfloor a^{b-1}\rfloor<n<\lfloor a^b\rfloor$. Couple that with the fact that $\lfloor a^b\rfloor$ is an increasing sequence, and you are done, because $$\forall b: \lfloor a^b\rfloor\neq n$$