Pablo Solis asked this at a recent 20 questions seminar at Berkeley. Is there a positive integer $N$, not of the form $10^k$, such that the digits of $N^2$ are all 0's and 1's?

It seems very unlikely, but I don't have a proof. It's easy to see that such a number must end in 1 or 9, and then easy to see that it must end in 01, 49, 51 or 99, and you can continue recursively for as long as you like, determining possible "suffixes". Using this, I had a computer check for me that there are no such N up to about $10^{24}$.

If you pretend that the digits of $N^2$ are randomly distributed, and $N$ has $n$-digits, there's a $(2/10)^{2n}$ chance of satisfying this condition. There are only $10^n$ $n$-digit numbers, so you might expect a $(4/10)^n$ chance of having a some $n$-digit number. This suggests we shouldn't expect to find anything.

(If you try the same problem in other bases, where the probabilities are better, you do find a few: in base 5, 222112144, 22222111221444 and 100024441003001 work.)