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Is $357911$ the only perfect power that is obtained by concatenating $5$ or more consecutive odd numbers(in decimal)?

I noticed that while memorizing all perfect cubes from $1$ to $10^{6}$, $357911=71^{3}$ is a perfect cube that is obtained by concatenating $5$ consecutive odd numbers(namely $3, 5, 7, 9,$ and $11$).

But is $357911$ the only perfect power is obtained by concatenating $5$ or more consecutive odd numbers?

I know these things:

  • If a odd number is a perfect square, then the number must be $1\pmod{8}$, so if we want to make a perfect square that is obtained by concatenating $n$ odd numbers, then the number must be (either) form of $40k+1$ or $40k+9$.
  • If we want our number that we concatenated to be a perfect 5th power, then the last two digits must either one of these:$01, 07, 25, 43, 49, 51, 57, 75, 93, 99$.

I have no idea for other perfect odd prime powers.

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  • $\begingroup$ For numbers ending in $5$, all their powers (except the zeroth and first powers) must end with either $25$ or $75$ (as they are odd numbers divisible by $25$), and of course, only numbers ending in $5$ could have powers ending in $5$. This places severe restrictions for numbers ending in $5$. $\endgroup$ Feb 9 at 5:09
  • $\begingroup$ How far have you checked with brute force ? $\endgroup$
    – Peter
    Mar 15 at 11:53
  • $\begingroup$ After having checked some larger ranges , I arrived at the conjecture that $357911$ is the only one. $\endgroup$
    – Peter
    Mar 15 at 13:21

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