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For an integer $m \in (\sqrt {10} , 10)$ , define $A_{10,m}:=\{n \in \mathbb N : m^n=\sum_{j=0}^k 10^j m^{n_j} ; n_j=0 $ or $1; k \ge 0\}$ . So , $A_{10,m}$ is the set of those natural numbers , raised to which power of $m$ , its digits in base $10$ representation also consists of powers of $m$ only . Then can we say that for which such integers $m \in (\sqrt {10} , 10)$ , is $A_{10,m}$ a finite set ? In general what can we say if $10$ is replaced by some $d \ge 3$ ?

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    $\begingroup$ Do you know any examples containing an $n>1$? $\endgroup$ Commented Oct 28, 2017 at 18:54
  • $\begingroup$ @ZackWolske : In base $10$ ? No, I don't ... though I have checked only upto $n=10$ ... and I have not checked in any other base $d$ $\endgroup$
    – user111524
    Commented Oct 29, 2017 at 6:38

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