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.
Just some working:
if $x=d_m d_{m-1} d_{m-2}\dots d_1 d_0, \space d_0 = 1, d_i=0 \text{ or } 1$ is the number we're looking for,
denote $x_n=d_m d_{m-1} d_{m-2}\dots d_{n+1} d_n$
So, $$ x =x_0 = 10x_1+d_0 = 10x_1+1 =y^2 $$ for some $y$ $$ 10x_1+1=y^2 $$ $$ 10x_1=y^2-1 $$ $$ 10x_1=(y+1)(y-1) $$ $$ 2.5.x_1=(y+1)(y-1) $$
so $$ y+1=2k \text{ or } 5k $$ or $$ y-1=2k \text{ or } 5k $$
$$ y=2k-1 \text{ or } 5k-1 \text{ or } 2k+1 \text{ or } 5k+1 $$
I.e $ y = 1 \mod 2, \text{ or } \space 1 \text{ or } 4 \mod 5 $.
Also $ x_1=10x_2 + d_1 $ so $ x_1=10x_2 \text{ or } x_1=10x_2 + 1 $
I.e. $$ 10(10x_2[+1])+1=y^2 $$ $$ 100.x_2=y^2-[1,11] $$ So $$ 10^2.x_2=[2k-1 \text{ or } 5k-1 \text{ or } 2k+1 \text{ or } 5k+1]^2-[1,11] $$ $$ =2^2k^2-2k-[0,10] $$ or $$ =2^2k^2+2k-[0,10] $$ or $$ =5^2k^2-10k-[0,10] $$ or $$ =5^2k^2+10k-[0,10] $$
Similarly $ x_2 = 10x_3+d_2 = 10x_3 \text{ or } 10x_3 + 1 $
$$ 10^3.x_3 $$ $$ =2^2k^2-2k-[0,10,20] $$ or $$ =2^2k^2+2k-[0,10,20] $$ or $$ =5^2k^2-10k-[0,10,20] $$ or $$ =5^2k^2+10k-[0,10,20] $$
Generalising $$ 10^n.x_n $$ $$ =2^2k^2-2k-10l $$ or $$ =2^2k^2+2k-10l $$ or $$ =5^2k^2-10k-10l $$ or $$ =5^2k^2+10k-10l, 0<=l<n $$
So firstly, $ \mod 10 $ $$ 2^2k^2-2k = 0 $$ or $$ 2^2k^2+2k = 0 $$ or $$ 5^2k^2 = 0 $$
i.e $$ k = 0 \mod 2 $$
let $ 2j=k $ i.e. $$ 10^n.x_n $$ $$ =16j^2-4j-10l $$ or $$ =16j^2+4j-10l $$ or $$ =100j^2-20j-10l $$ or $$ =100j^2+20j-10l, 0<=l<n $$
And for anyone looking on, the $j$, $l$ and which "or" formulae can (will) vary for each $n$.