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

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.

• 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 $5$. Feb 9 at 5:09
• How far have you checked with brute force ? Mar 15 at 11:53
• After having checked some larger ranges , I arrived at the conjecture that $357911$ is the only one. Mar 15 at 13:21