TITLE: Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square?
QUESTION [21 upvotes]: 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.

REPLY [7 votes]: 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 $3 333 333 333<\sqrt n\;<3 333 333 334.$