Are there any perfect squares of the form 88...81 (in decimal, at least two 8's)? I saw this problem recently and it is deceptively hard.  The usual mod 4 trick won't work, and indeed there will be perfect squares whose last n digits will be 88...81, for any n.
I can show that if 88...81 is a perfect square, $2n$ digits total, then the decimal representation of $\sqrt{8/9}$ must start with $n$ digits followed by almost that many zeroes (like $n-1$ or so).  This seems preposterous, but I can't figure out how to rule it out.  I've thought of using continued fractions, since there is a limit on how well $\sqrt{8/9}$ can be approximated with rationals, but I can't seem to get a tight enough inequality.  Maybe someone with more expertise in Diophantine approximation can answer this?
Note: even if you can prove that there aren't infinitely many perfect squares of this form, even if you don't have a computable bound on how large they might be, I'd still be interested.
 A: The number $88\cdots81$ in general form is
$$
(10^n - 1){8 \over 9} - 7 \text{ where } n \text{ is the total number of digits.}
$$
If we need this to be a perfect square, we need to solve the Diophantine equation
$$
(10^n - 1){8 \over 9} - 7 = y^2, y \in \mathbb{Z}
$$
This implies
$$
y = \pm \bigg({1 \over 3} \sqrt{2^{n + 3} 5^{n} - 71}\bigg)
$$
The term under the radical needs to be a perfect square for $y$ to be an integer and also a multiple of $9$. $\text{i.e.,} \exists x \in \mathbb{Z}$ such that,
$$
2^{n + 3} 5^{n} - 71 = (3x)^2
$$
Case 1 ($n$ is even):
Since $n$ is even, we may take $n = 2m$ and rewrite the equation as
$$
2^3(10)^{n} - 71 = (3x)^2 \\
\implies 2^3(10)^{2m} - 71 = (3x)^2 \\
\implies 8(10^m)^{2} - 71 = (3x)^2
$$
Taking $10^m = Y, 3x = X$,
$$
8Y^{2} - X^2 = 71
$$
This is a Pell equation.
Case 2 ($n$ is odd):
Since $n$ is odd, we may take $n = 2m+1$ and rewrite the equation as
$$
2^3(10)^{n} - 71 = (3x)^2 \\
\implies 2^3(10)^{2m+1} - 71 = (3x)^2 \\
\implies 80(10)^{2m} - 71 = (3x)^2 \\
\implies 80(10^m)^{2} - 71 = (3x)^2
$$
Taking $10^m = Y, 3x = X$,
$$
80Y^{2} - X^2 = 71
$$
This is also a Pell equation.
We may solve those Pell equations to find solutions for $n$.
