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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.

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  • $\begingroup$ By "rep-unit" do you mean to case $k=1$? $\endgroup$
    – Vincent
    Jan 6, 2019 at 15:49
  • $\begingroup$ @Vincent Yes, a repunit has the form $1\cdots 1$ $\endgroup$
    – Peter
    Jan 6, 2019 at 15:49

1 Answer 1

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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.

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  • $\begingroup$ And the result is that there are no perfect powers except $4,8,9$ ? $\endgroup$
    – Peter
    Jan 7, 2019 at 18:39
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    $\begingroup$ 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 $7$. $\endgroup$
    – Mike Bennett
    Jan 7, 2019 at 23:47
  • $\begingroup$ Amazing, thank you very much, Mike! I only wonder why $1$ is considered to be a perfect power. $\endgroup$
    – Peter
    Jan 7, 2019 at 23:58

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