Can $36\underbrace{11…1}_{n \text { times}}$ be a perfect cube?
Inspired from the question
What is the next perfect square of the form 14444… in decimal notation? ,$361$ is the only perfect square that is form of $361…1$, since the others is $3\pmod{4}$, but can $36\underbrace{11…1}_{n \text{ times}}$ be a perfect cube when $n\geq2$?
I tried to use Pari GP and checked the values of $n\leq10^4$ manually, but none of the numbers shown is a perfect cube.
Hint: $36\underbrace{11…1}_{n \text{ times}}\equiv3\pmod{4}$ when $n\ge2$ and $36\underbrace{11…1}_{n \text{ times}}\equiv0,\pm1\pmod{9}$ when $n\equiv0,\pm1\pmod{9}$.
Best Answer
Let $N=3611...1$. We have $9N=325\cdot 10^n-1\equiv -1 \pmod {13}$, thus $N \equiv 10 \pmod {13}$. However $10$ is not a cubic residue mod $13$, hence $N$ cannot be a perfect cube.