Is exactly one of $a, a + b^n, ..., a + b^{p - 1}$ divisible by $p$? Consider the following structure over $\mathbb{N_0}$: $2, 2 + 3^n, 2 + 3^{n + 1}, ..., 2 + 3^{n + 3}$. Here we have five numbers and it's not hard to prove that exactly one if them is divisible by $5$.
What about general result?
Consider a general structure over $\mathbb{N_0}$: $a, b, p \in \mathbb{N}$):
$a, a + b^n, a + b^{n + 1}, ..., a + b^{n + p - 2}.$ Of course, assume $a \not\equiv 0 \pmod p.$
My hypothesis is that exactly one of $a + b^i, i = n, n + 1, ..., n + p - 2$ is divisible by $p$, but does this property holds for any $p, b$? Or maybe just if $(b, p) = 1$ or one of them (or both?) is a prime? How to prove this?
 A: The statement that exactly one of $\{a+s\colon s\in S\}$ is a multiple of $p$ for every $a$ given a set $S$ of size $p$ is equivalent to $S$ having one element with every remainder modulo $p$. (For an element $s\in S$ of remainder $r$, the number $a+s$ will be a multiple of $p$ for every $a\equiv -r\bmod p$, and this covers every $a$ exactly once as you vary $r$.)
Your set $S$ consists of $b^n$ times every element of $\{0,1,b,b^2,\dots,b^{p-2}\}$. If $b$ is a multiple of $p$ this clearly isn't going to work, so $S$ consisting of all residues modulo $p$ is the same as $\{0,1,b,b^2,\dots,b^{p-2}\}$ consisting of all residues modulo $p$. It is clear that this set has only one zero mod $p$, so it's the same as powers of $b$ reaching every nonzero value modulo $p$.
Unfortunately given a pair $(b,p)$ it isn't obvious whether it satisfies this property. However, what you do know is that $b^k$ can't be $1$ modulo $p$ for any $1\leq k\leq p-2$.  However, by the pigeonhole principle, the numbers $\{b^0,b^1,\dots,b^{p-1}\}$, as there are $p$ of them and none are $0$ modulo $p$, must contain some two elements that are equivalent modulo $p$. If $b^i\equiv b^j\bmod p$, then $b^{i-j}\equiv 1$, and the only way for this to happen while keeping all residues from $b^0$ to $b^{p-2}$ distinct is if $b^{p-1}=1\bmod p$ but no $b^k$ with $0<k<p-1$ is. In fact, $b^{p-1}\equiv 1\bmod p$ for all $b$ (this is Fermat's little theorem), but you can show via an extension of this pigeonhole argument that the $b$ that satisfy your condition are exactly those for which $p-1$ is the minimal such exponent. As noted in the comments, these are called primitive roots, and their existence for every prime $p$ is a rather important theorem of number theory.
