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I don't think a number whose decimal expansion looks like 10010111010110001 can be a square number. However, I just can't prove it.

More precisely, does there exist a positive square integer, whose decimal expansion consists only of $0$s and $1$s, apart from the obvious $100^n$, $n\ge 0$?

And if it exists, can it be a power of 3?

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    $\begingroup$ 0, and $10^{2n}$ are square numbers whose digits are 0s and 1s $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Mar 21, 2022 at 7:39
  • $\begingroup$ You have asked n the wrong forum. $\endgroup$
    – Gerald Edgar
    Commented Mar 21, 2022 at 9:24
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    $\begingroup$ i think in base 2 it is true $\endgroup$
    – Pietro Majer
    Commented Mar 21, 2022 at 10:41
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    $\begingroup$ I've edited the question, which seems reasonable to me here. Indeed heuristics indicate that in base $p$ there should be only finitely many solutions: there are $2^n$ candidates in the $\sim p^n$ numbers between $p^n$ and $p^{n+1}$ to have such an expansion, and there are $p^{n/2}$ squares therein. Assuming $p\ge 5$, the probability that some given $(1,0)$-number being a square is $p^{-n/2}$ so the probability that at least one being a square should be $\le (2/\sqrt{p})^n$ and hence, summing, the probability to have a solution $\ge p^n$ should be in $O((2/\sqrt{p})^n)$ too. $\endgroup$
    – YCor
    Commented Mar 22, 2022 at 8:20
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    $\begingroup$ More precisely, this is a duplicate of mathoverflow.net/questions/22/… $\endgroup$
    – YCor
    Commented Mar 22, 2022 at 10:42

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