Let $M$ be the set of positive integers having no consecutive equal digits in the decimal expansions, in other words the strings $00$ , $11$ , $\cdots$ , $99$ do not occur.
Does $M$ contain infinite many squares ? What about cubes, fourth-powers and so on ?
I found a large example : The square of the $135$-digit number $$61987608238664627631443019969085317171771617953023827580498834952037$$ $$0593591927886368167280334245269806736293723763536153215722202056806$$
has $270$ digits and is a member of $M$.The idea was to start with a small number such that its square is in $M$ and if this number (denote if with $m$) had $n$ digits, I searched a number of a form $k\cdot 10^n+m$ having again the property that its square is in $M$. Since the last digits do not change, there is a "good" chance that we find such a number relatively early.
However, this method is also limited and each further number took more and more time. I think I must find a pattern that ensures infinite many squares in $M$ (assuming that there are infinite many).
Finally, it seems certain that $M$ contains infinite many primes, but can this be proven ?
$(\frac{10^{27}-1}{9})^2 = 12345679012345679012345678987654320987654320987654321$
More generally $(\frac{10^{9k}-1}{9})^2$ gets you $k-1$ $\overline{123456790}$ blocks, one $\overline{12345678}$ block, $k-1$ $\overline{987654320}$ blocks and a $\overline{987654321}$ block.
A simple way to prove this is noticing $(\frac{10^{9k}-1}{9})^2 = \frac{10^{18k}-1}{81} - 2 \frac{10^{9k}-1}{81}$, and $\frac{10^{9m}-1}{81}$ is just $\lfloor \frac{10^{9m}}{81} \rfloor$, so you get something that has to do with the decimal expansion of $\frac{1}{81} = 0.012345679...$.
Some formulas for third powers might come from the expansion of $\frac{1}{729}$