Prove that if $\gcd(n, m) = 1$, then $\gcd(R_n, R_m) = 1$

Question

We will call a number with $i$ consecutive $1$s in its decimal representation a repunit and denote it by $R_i$.

Prove that if $\gcd(n, m) = 1$, then $\gcd(R_n, R_m) = 1$.

This looked like a proof by contradiction, so I assumed that $\gcd(n, m) = 1$ and $\gcd(R_n, R_m) = d \neq 1$. We are required to find a contradiction.

Now, the above implies that for some integers $a$ and $b$, $da = R_n$, and $db = R_m$. Assume without loss of generality that $R_n \leq R_m$. This means $R_m = R_n\cdot 10^{m-n} + R_{m - n}$

Clearly, $d \mid R_n\cdot 10^{m-n} \implies d\mid R_{m-n}$

Also, $R_{m+n} = R_m\cdot 10^n + R_n = db\cdot 10^n + da = d(10^nb + a) \implies d\mid R_{m+n}$

Even after numerous manipulations of the above results I could not produce a contradiction.

Answer 1

HINT:

If $$R_{(n,b)}=\underbrace{(11\cdot11)_b}_{n \text{ digits}}=\frac{b^n-1}{b-1}$$ where base $b$ is a positive integer $>1$

Using this, $$(b^n-1,b^m-1)=b^{(n,m)}-1$$ where $m,n$ are positive integers

$$\implies\left(R_{(n,b)},R_{(m,b)}\right)=R_{((n,m),b)}$$

Answer 2

Let $gcd(m,n)=d$
Then we have $$mx+ny=d$$ for $x,y\in \mathbb Z$

Let , $gcd(10^{m}-1,10^{n}-1)=e$, then
$$10^m\equiv 1 \mod e$$ $$10^n\equiv 1 \mod e.$$

So, we have
$$10^{mx+ny}\equiv1 \mod e$$ from which follows
$$10^d\equiv 1 \mod e$$ from which follows
$$e|10^{d}-1\tag1$$

Since $d=gcd(m,n)$
we have $$d|m$$ $$d|n.$$ Thus, for some $k,l$,
$$m=dk$$ $$n=dl.$$
Now, for some $g,h$
$$10^{m}-1=(10^d)^{k}-1=(10^{d}-1)g$$ $$10^{n}-1=(10^d)^{l}-1=(10^{d}-1)h.$$ Thus
$$(10^{d}-1)|(10^{m}-1)$$ $$(10^{d}-1)|(10^{n}-1).$$
So by the definition of GCD we have

$$(10^{d}-1)|e\tag2$$
From (1) and (2) we have

$$e=(10^{d}-1)$$ $$gcd(10^{m}-1,10^{n}-1)=(10^{d}-1)$$ in form $gcd(m,n)=d$.

Now, as $gcd(m,n)=1$ we have

$$gcd(10^{m}-1,10^{n}-1)=10^1-1=9$$ thus
$$gcd((\frac{10^{m}-1}{9}),(\frac{10^{n}-1}{9}))=1.$$
$$gcd(R_m,R_n)=1.$$