Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square? Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square?
I haven't been able to figure out which properties of perfect squares disallows such a number from being a perfect square. 
 A: If $|a|<1$ then $1-\frac {|a|}{2+|a|}\leq \sqrt {1+a}\;\leq 1+\frac {|a|}{2}.$
If $n$ is a $20$-digit number beginning with $11$ ones then $n=(1+a)10^{20}/9$ with $|a|<10^{-10},$ so $$\left(1-\frac {|a|}{2+|a|}\right)10^{10}/3 \leq\sqrt n\;\leq \left(1+\frac {|a|}{2}\right)10^{10}/3$$ which immediately implies $3 333 333 333<\sqrt n\;<3 333 333 334.$
A: That's because\begin{align*}3\,333\,333\,333^2&<11\,111\,111\,111\,000\,000\,000\\&<11\,111\,111\,111\,999\,999\,999\\&<3\,333\,333\,334^2.\end{align*}
A: There is no integer between $\sqrt{11111111111000000000}$ and $\sqrt{ 11111111112000000000}$
A: Let's show more generally that a $2n$-digit square cannot begin with $n+1$ $1$'s.
Note that the smallest $2n$-digit number beginning with $n+1$ $1$'s is
$$11{\ldots}110{\ldots}0={10^{2n}-10^{n-1}\over9}$$
and the largest is
$$11{\ldots}119{\ldots}9={10^{2n}-10^{n-1}\over9}+10^{n-1}-1={10^{2n}+8\cdot10^{n-1}-9\over9}$$
Since $10^{n-1}\lt2\cdot10^n-1$ while $8\cdot10^{n-1}-9\lt2\cdot10^n+1$, we obtain
$$\left(10^n-1\over3 \right)^2={10^{2n}-2\cdot10^n+1\over9}\lt{10^{2n}-10^{n-1}\over9}=11{\ldots}110{\ldots}0$$
while
$$11{\ldots}119{\ldots}9={10^{2n}+8\cdot10^{n-1}-9\over9}\lt{10^{2n}+2\cdot10^n+1\over9}=\left(10^n+1\over3\right)^2$$
Thus if $11{\ldots}110{\ldots}0\le N\le11{\ldots}119{\ldots}9$, then
$${10^n-1\over3}\lt\sqrt N\lt{10^n+1\over3}\lt{10^n+2\over3}$$
But $(10^n-1)/3$ and $(10^n+2)/3$ are consecutive integers (their difference is $1$).  So $\sqrt N$ cannot be an integer.
Remark 1:  This answer owes much of its logic to a now-deleted answer by achille hui.
Remark 2: This answer was motivated by JollyJoker's first comment beneath the OP.  JJ's follow-up comment there makes it clear that squares with $2n+1$ digits can begin with more than $n+1$ $1$'s.
A: Let $x$ be a positive integer with $2n$ digits when written in base $b>1$, where the $n+1$ first digits are 1's, and where we will assume that $b-1=d^2$ is a perfect square. Note that this holds in the decimal system ($b=10$), as $10-1=9=3^2$. It also holds in the binary system ($b=2$), and with respect to several other bases.
Assume that $x$ is a perfect square, $x=c^2$. Then
$$b^{2n-1}+\dots+b^{n-1}\leq c^2\leq b^{2n-1}+\dots +b^{n-1}+(b-1)\Bigl(b^{n-2}+\dots+1\Bigr)$$
Summing the geometric series, we get
$$b^{n-1}{b^{n+1}-1\over b-1}\leq c^2\leq b^{n-1}{b^{n+1}-1\over b-1}+(b^{n-1}-1)$$
Multiply by $b-1=d^2$ and subtract $b^{2n}$. This gives
$$-b^{n-1}\leq c^2d^2-b^{2n}\leq -b^{n-1}+(b-1)(b^{n-1}-1)$$
It follows that
$$-b^{n-1}\leq (cd+b^n)(cd-b^n)<b^{n}$$
Now $d$ divides $b-1$, and hence $d$ cannot divide $b^n$. So $cd-b^n$ is a nonzero integer, but then one of the inequalities above must be violated. 
It follows that the assumption that $x$ was a perfect square cannot hold.
Edit: An example to show that the result is not true in all bases. Let $b=n=3$, and observe that $(111101)_3=1\cdot 3^5+1\cdot 3^4+1\cdot 3^3+1\cdot 3^2+1\cdot 3^0=361=19^2$ is a perfect square with $n+1=4$ initial 1's. But then $b-1=2$ is not a perfect square, as was required in the argument above.
