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Let us consider $R_n:=\frac{{10}^n-1}{9}=11\cdots1$. Let $m\ge 2\in\mathbb N$.

Question : Do there exist natural number solutions such that $x^m=R_n$ for $m$ such that $(m,10)=1$ ?

Here, suppose that $(a,b)$ represents the greatest common divisor of $a,b$.

Motivation : I've known the following theorem :

Theorem : There does not exist any natural number solution such that $x^m=R_n$ for $m$ such that $(m,10)\not=1$.

We can easily prove this theorem by considering the $m=2,5$ cases. Then, I've thought about the other $m$, but I'm facing difficulty. Can anyone help?

In the following, I'm going to show what I've got.

$(1)$ : $R_{pq}$ is a multiple of $R_p$ (If $n$ is not a prime number, then $R_n$ is not a prime number).

$(2)$ : Prime factorization for $R_n$ for prime numbers $3\le n\le 17$ (because of $(1)$).

$$R_3=3\cdot 37, R_5=41\cdot 271, R_7=239\cdot 4649, R_{11}=21649\cdot 513239,$$$$ R_{13}=53\cdot 79\cdot 265371653, R_{17}=2071723\cdot 5363222357.$$

I also noticed that each of $R_{19}$ and $R_{23}$ is a prime number.

$(3)$ : $(R_n,R_m)=R_{(n,m)}.$ (We can prove this by the Eucledean algorithm.)

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up vote 1 down vote accepted

This result is a special case of a theorem of Bugeaud and Mignotte (see "Sur l'équation diophantienne (x^n−1)/(x−1)=y^q. II", C. R. Acad. Sci. Paris Sér. I Math. 328 (1999), no. 9, 741--744). The proof is not elementary (it uses lower bounds for linear forms in $p$-adic logarithms).

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