# Condition for Repunits to be Coprime

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## Theorem

Let $R_p$ and $R_q$ be repunit numbers with $p$ and $q$ digits respectively.

Then $R_p$ and $R_q$ are coprime if and only if $p$ and $q$ are coprime.

## Proof

### Necessary Condition

Let $R_p$ and $R_q$ be coprime.

Aiming for a contradiction, suppose $p$ and $q$ are not coprime.

Let $p = d m, q = d n$ for some $d > 1$.

$R_d \divides R_p$

and:

$R_d \divides R_q$

where $\divides$ denotes divisibility.

That is, $R_d$ is a common divisor of $R_p$ and $R_q$.

This contradicts our hypothesis that $R_p$ and $R_q$ are coprime.

Thus by Proof by Contradiction, $p$ and $q$ are coprime.

$\Box$

### Sufficient Condition

Let $p$ and $q$ be coprime.

Let $R_p$ and $R_q$ be repunits in base $b$.

By Integer Combination of Coprime Integers, there exists integers $m, n$ such that $m p + n q = 1$.

One of $m, n$ is strictly positive while the other is negative.

Without loss of generality, suppose $m > 0$ and $n \le 0$.

Then from $m p = 1 - n q$:

We have $R_{m p} - b R_{- n q} = 1$.

By Divisors of Repunit with Composite Index, $R_p$ is a divisor of $R_{m p}$, while $R_q$ is a divisor of $R_{- n q}$.

By definition of divisibility, there exists integers $c, d$ such that $R_{m p} = c R_p, R_{- n q} = d R_q$.

Therefore $c R_p - b d R_q = 1$.

Note that both $c$ and $- b d$ are integers.

By Integer Combination of Coprime Integers, $R_p$ and $R_q$ are coprime.

$\blacksquare$