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 ?