Base Ten Digit Orientation of Perfect Squares Recently I encountered a problem involving perfect squares. After struggling with it for some time, I came upon  the realization that there are no perfect squares with digit orientation (in base ten) $AABB$, $AAABBB$, $AAAABBBB$, and so on. This seems to make sense intuitively, but I don't know how to prove it. If it is true, it definitely helps my solution.
 A: Let us consider $AABB$ in detail.  It clearly has a factor $11$.  To be a perfect square, it therefore must have a factor $11^2=121$.  The multiplier of $121$ must also be a perfect square and be at least $9$ to get four digits.  $9$ doesn't work, giving $1089$.  As the ones digit of the square is $B$ and the ones digit of $121$ is $1$ the ones digit of the square must be $B$, so you are wanting $CB \cdot 121=AABB$ with $CB$ a square.  The tens digit of $AABB$ comes from the ones digit of $C\cdot 1 + B \cdot 2$ which says that $C+B=10$.  The only two digit square whose digits add to $10$ is $64$ and indeed $64\cdot 121=7744=88^2$
A: $\underbrace{AAA...}_{n}\underbrace{BBB...}_n=A*\underbrace{1111...}_n*10^n + B*\underbrace{11111...}_n =\underbrace{1111....}_n(A*10^n+ B)$
which can only be a perfect square if $ \underbrace{111...}_n$ is.
Which would require there to be some $N = 10a + b$ so that $N^2 \equiv 11 \mod 100$.  That would require $b^2 \equiv 1 \mod 10$ so $b = 1,9$ so $N \equiv 10c \pm 1\mod 100$ and $N^2 \equiv \pm 20c + 1\equiv 11\mod 100$.  This requires $-2c \equiv 1 \mod 10$ which is, of course, impossible.
