Odd perfect squares whose decimal representation consist only of 1's and o's Are there any odd perfect squares (apart from the trivial 1), whose decimal representations only uses 1 and 0?
Working modulo 8, we can get that the last 3 digits must be 001. However, since $4251^2 = 18071001 $, there goes my hope of showing that the last $n$ numbers be be of the form $0 0 \ldots 0 0 1$.

This question is motivated by the standard questions of asking if the repunit can be a perfect square, or if $10^n+1$ can be a perfect square.
 A: Edit: Yep, as pointed out in the comments, my proof is wrong. I totally expected something like this. The fatal problem is that I missed some counterexamples in the general case.
I'm leaving this answer up, because as a rule I don't delete posts just for stuff like this, but that's what you get for trying to dive in head-first, I guess.
/edit

I know this is an old question, but I figured this would be a good exercise. First post on MSE, yadda yadda yadda. Hopefully the proof checks out.
I claim that there are no odd integers $x > 1$ such that the decimal representation of $x^2$ contains only the digits 0 and 1.
Suppose $x^2\equiv 1 \mod 10$. Then
$$x \equiv \pm1 \mod 10$$
Now we have $$x^2\equiv 1\vee x^2\equiv 11 \mod 100$$
($81=9^2$ is not counted since we are only allowed to use the digits 0 and 1).
But $11$ is not a perfect square. So that leaves only
$$x^2 \equiv 1 \mod 100.$$
In general, we can suppose we have proven that
$$x^2\equiv 1 \mod 10^k$$
It follows from this that
$$x\equiv \pm 1 \mod 10^k$$
Now notice that
$$x^2\equiv 1 \vee x^2\equiv 10^{k}+1 \mod 10^{k+1}$$
since by incrementing $k$, we are essentially adding another digit to the left of the number. By our rule, this extra digit can only be 0 or 1.
Now we need to show that $\sqrt{10^k+1}$ is irrational (see below). Then we are left with $$x^2\equiv 1 \mod 10^{k+1}.$$
By induction, therefore, $x^2\equiv1 \mod 10^k$ for all $k \ge 1$. Thus, the trivial $x^2=1$ is the only solution.

Now for my argument that $\sqrt{10^k+1}$ is irrational for all $k$:
Suppose $k\ge 0$ is even; then $10^k$ is a perfect square; thus $10^k+1$ is not a perfect square.
The case for odd $k$ can be verified numerically: we simply note that $\sqrt{10^k+1}-\sqrt{10^k}$ converges to zero for large $k$, and that $\sqrt{10^k}$ is irrational for odd $k$.
