Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question

1 Answer 1

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).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.