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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?

J. W. Tanner
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1b3b
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    $3$ is a primitive root $\pmod 5$, that's what's going on here. Try $4$ instead of $3$. – lulu Oct 05 '20 at 20:02
  • Unfortunately, I haven't studied about primitive roots yet, can you please be more concrete? – 1b3b Oct 05 '20 at 20:05
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    Sure. Every number prime to $5$ is congruent to one of $3, 3^2, 3^3, 3^4\pmod 5$ That's what "primitive root" means. If you want to stick with $b=3$, try $p=11$ instead. This time we get $3^5\equiv 1 \pmod {11}$, so you either miss a value or you get it twice in your list. – lulu Oct 05 '20 at 20:08
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    To stress: you don't need to know what a primitive root is to check the counterexamples I have proposed. For the first one, I suggest taking $b=4, p=5, n=1$. For the second, I propose $b=3, p=11, n=1$. – lulu Oct 05 '20 at 20:09
  • One intermezzo: how to prove $0, a^n, a^{n + 1}, ..., a^{n + b - 2}$ is complete residue system $\text{mod} \ b$ for any natural $a, b$ such that $(a, b) = 1$? – 1b3b Oct 05 '20 at 20:20
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    Again, it's not true in general. The same counterexamples apply. Try $a=3, b=11, n=1$. Then your list is ${0,3,9,5,4,1,3,9,5,4,1}\pmod {11}$. – lulu Oct 05 '20 at 20:24
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    The Primitive Root Theorem I alluded too earlier says that, for any prime $p$, there is a residue $g$ such that ${1, g, g^2, \cdots, g^{p-1}}$ is a complete list of non-zero residues $\pmod p$. But this is not true for all $g$ nor is it true if you replace $p$ by a general composite (though there are composites for which it is true). – lulu Oct 05 '20 at 20:27
  • Aha... I understand now that part. So, in the last example, $b$ has to be primitive root of $a$? – 1b3b Oct 05 '20 at 20:27
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    Well, $a$ has to be a primitive root for $b$. Thus $3$ is a primitive root for $5$ but not for $11$. – lulu Oct 05 '20 at 20:28
  • So condition $a$ has ti be primitive root of $b$ a necessary and sufficient condition? – 1b3b Oct 05 '20 at 20:29
  • Let us continue this discussion in chat. – 1b3b Oct 05 '20 at 20:29
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    Sorry, I can't talk further right now. Good luck! – lulu Oct 05 '20 at 20:30
  • Okay, thank you! – 1b3b Oct 05 '20 at 20:31

1 Answers1

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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.

Carl Schildkraut
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