<s>[INST] 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. 
 [/INST] 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.
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