perfect square numbers with $0$ and $1$ I want to show that there is not an integer with digits of only $0$ and $1$ that has at least two $1$ and is a complete square number.
I tried to prove it by induction, but I couldn't.
 A: Not an answer, just a bunch of necessary conditions. So suposse that $N$ has only zeros and ones, is a square but not a power of $10$:


*

*If there exists such a number, then there exists one that ends with $1$. Indeed, if $N$ ends with zero, then it must have an even number of trailing zeros. If you remove them, the result must be a perfect square. From now, assume that $N$ ends with $1$.

*$\sqrt N$ ends with $1$ or $9$, and begins with $3$ or $1$.

*The number of ones, $k$, is congruent to $N$ modulo $9$. Then $k$ must be a square modulo $9$. That is $k\equiv 0,1,4\text{ or }7\pmod 9$.

*The penultimate digit is $0$, because $11$ is not a square modulo $100$.

*$\sqrt N\equiv \pm1\pmod{50}$ (thanks to Thomas Andrews).


I'll edit this "answer" if I find more of them.
An heuristic reasoning: the probability that a number of at most $n$ digits has only ones and zeros is $1/5^n$. And that it is a perfect square is $1/10^{n/2}$. Assuming that both events are independent, the probability that both conditions hold for the same number is $1/(5\sqrt{10})^n$, which is lesser than $1/10^n$. So... probably, the answer is no.
