Prime dividing repunit

Question

Let $ R(n) = \underbrace{111\ldots111}_{\text n\ ones}$. Prove that if a prime number $ p \neq 3 $ divides $ R(n) $ then $ n $ and $ p - 1 $ are not coprime.

So obviously $ R(n) = \frac{10^n - 1}{9}$. Now if $ p $ divides $ R(n) $ then

$$ \frac{10^n - 1}{9} \equiv 0 \pmod p $$

which implies

$$ 10^n - 1 \equiv 0 \pmod p $$

$$ 10^n \equiv 1 \pmod p $$

Can we deduce from there that $ GCD(n, p - 1) \neq 1 $? How? Also, why is $ p \neq 3 $ requirement necessary? I suppose "multiplying both sides" by $ 9 = 3^2 $ is somehow relevant, but I'm not sure why... it's not diffucult to come up with a counterexample for $ p = 3 $ case, but I don't know how the proof would account for it.

Answer 1

Use little Fermat $10^{p-1} \equiv 1\bmod p$

Answer 2

Theorem(1): Let $m$ and $a$ be two integers; such that $\gcd(a,m)=1$.
Then there exists a natural number $r$ such that $a^{r} \overset{m}{\equiv} 1$.
The least positive integer $r$ by the above property will be called the order of $a$ module $m$;
and we will denote it by $\text{ord}_m(a)$.

Theorem(2): Let $m$ and $a$ be two integers; such that $\gcd(a,m)=1$.
Suppose that $a^R \overset{m}{\equiv} 1$; then we can conclude that: $$\text{ord}_m(a) \mid R \ \ .$$

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Remark(I): Let $p\neq3$ (also $p\neq2$ and $p\neq5$ ) be a prime number; then $1 < \text{ord}_p(10)$.
proof: Suppose on contrary that $ \text{ord}_p(10) = 1$; then we must have:
$$ 10^1 \overset{p}{\equiv} 1 \Longleftrightarrow p \mid 10-1 \Longleftrightarrow p =3; $$ which has an obvious contradiction with our assumption.

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Remark(II): By little Fermat's theorem we know that $$10^{p-1} \overset{p}{\equiv} 1 ;$$ now Theorem(2) implies that $\text{ord}_p(10) \mid p-1$.

Remark(II): From the assuptuion of the question we know that $$10^{n} \overset{p}{\equiv} 1 ;$$ now Theorem(2) implies that $\text{ord}_p(10) \mid n$.

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By the last two remarks we can conclude that:

$$1 < \text{ord}_p(10) \mid \text{gcd}(n, p-1).$$