Given that $$U_n=\underbrace{1\cdots1}_{n\text{ times}}$$ and $n >2$, how can one show that if $m$ and $n$ are co-prime, then $U_n$ and $U_m$ are also co-prime?
Because $U_m= \frac{10^{m}-1}{9}$ and $U_n= \frac{10^{n}-1}{9}$, follows that $U_{n} > 10 U_{m}$, so there is always a prime between them due to the Bertrand's postulate.
Can I prove it using an other way ?
This is an IMO exercise so I think I don't have to use Bertrand's prostulate ! Or if I use it I have to prove it
Edit 1:
I tried using Lemma
We have
$9U_n=10^n-1$
$9U_m=10^m-1$
$\delta=\gcd(U_m,U_n)$
And we have $\gcd(m,n)=1$
So $\delta=10^{\gcd(m,n)}-1$
Where $\delta=10-1=9$
Finnaly $\gcd(9U_m,9U_n)=9$
Then $\gcd(U_n,U_m)=1$
Is it right ?
Let $q$ be any positive integer, and for $n \ge 0$ set $$ U_{n} = \frac{q^{n} -1}{q-1}. $$ You want to prove that for $m, n \ge 0$ $$\tag{gcd's} \gcd(U_{n}, U_{m}) = U_{\gcd(n, m)}. $$
This follows from the elementary fact that $U_{n}$ divided by $U_{m}$, with $m > 0$, leaves as a remainder $U_{r}$, where $r$ is the remainder of the division of $n$ by $m$. If you employ Euclid's algorithm on $U_{n}, U_{m}$, you will get the formula (gcd's).
In fact if $n = m t + r$, with $0 \le r < m$, then $$ U_{n} = U_{m} \cdot (q^{n-m} + q^{n - 2m} + \dots + q^{n - tm}) + U_{r} $$