# How sure can we be that $\underbrace {k\cdots k}_{m\ k's}$ cannot be a perfect power?

Let $$k$$ and $$m$$ be integers , $$m\ge 2$$ and $$k\in [1,9]$$. Denote $$n=k\cdot \frac{10^m-1}{9}=\underbrace {k\cdots k}_{m\ k's}$$

As far as I know, it is unknown whether a rep-unit can be a cube.

How sure can we be that $$n$$ can never be a perfect power ?

Some cases are easy. For $$k=2,4,5,6,9$$ it can be shown that a perfect power is impossible. The case $$k=8$$ cannot give a perfect power if we assume that a rep-unit cannot be a cube. A square can be easily ruled out.

I have checked that for $$m\le 10^4$$, we do not have a perfect power.

• By "rep-unit" do you mean to case $k=1$? – Vincent Jan 6 '19 at 15:49
• @Vincent Yes, a repunit has the form $1\cdots 1$ – Peter Jan 6 '19 at 15:49

This problem was completely solved by Bugeaud and Mignotte (see http://irma.math.unistra.fr/~bugeaud/travaux/chiffresrev.ps) in a paper in Mathematika in 1999; see Theorem 2. Their approach is based upon bounds for linear forms in $$p$$-adic logarithms.
• And the result is that there are no perfect powers except $4,8,9$ ? – Peter Jan 7 '19 at 18:39
• More generally, if $2 \leq b \leq 10$ and $1 \leq a \leq b-1$, an integer $N$ with all its digits equal to $a$ in base $b$ is a perfect power only for $N \in \{ 1, 4, 8, 9 \}$, $N=11111$ in base $3$, $N=1111$ in base $7$ and $N=4444$ in base $7$. – Mike Bennett Jan 7 '19 at 23:47
• Amazing, thank you very much, Mike! I only wonder why $1$ is considered to be a perfect power. – Peter Jan 7 '19 at 23:58