I think $111..12$ can be a perfect cube, since $111..12$ is always divisible by $8$ and is sometimes $0$,$1$ or $8$ mod $9$. Also $111..12$ is sometimes $1$,$8$, or $12$ mod $13$.
Can $111..12$ be a perfect cube? If yes, are there any perfect cubes form of $111..12$?
Best Answer
Partial solution:
Let $a=111\dots112=b^3$. Then $b^3-1=(10^c-1)/9$ for some integer $c$. Rewrite this as $10^c=9b^3-8$. We split into two cases, depending on the parity of $c$.
Case 1. If $c$ is even, then $10^c$ is a square, and we have $d^2=9b^3-8$ (where $d$ has to be a power of $10$). Multiplying through by $81$, we get $$ y^2=x^3-648, $$ where $y=9d$ and $x=9b$. This is the Mordell equation $y^2=x^3-k$, with $k=648$. Solutions of Mordell equations are tabulated at https://hr.userweb.mwn.de/numb/mordell.html#tbl5 where we find there are six solutions, $$ (x,y)=(9,9),(18,72),(22,100),(54,396),(97,955),(1809,76941) $$ Only the first of these satisfies the condition that $y$ be nine times a power of ten, and $(x,y)=(9,9)$ doesn't lead to a cube of the requested form, so we get no solutions when $c$ is even.
Case 2. If $c$ is odd, then $10^c$ is ten times a square, and $10d^2=9b^3-8$. By manipulations similar to those in Case 1, this leads to the Mordell equation $$ y^2=x^3-648000 $$ Unfortunately, the tables don't go that high, so we'd actually have to put in some work to find all the solutions, if any, to this equation. Perhaps someone else will take up the problem from here.