What squares remain squares when their leading digits are incremented?

Omitting solutions that are obtained from smaller solutions by
multiplying by powers of 10, the squares of these numbers satisfy the
condition:

1. (105,145), (3375,4625), (14025,17225), (326625,454625),
    (10846875,14753125), (43708125,53948125), ...

2. (45,55), (144375,175625), (463171875,560828125), ...

7. (2824483699753370361328125,2996282391593370361328125), ...

Here is how to find them. We have (y+x)*(y-x) = 10^n, and so we must have 
{y+x, y-x} as {5^m*10^a, 2^m*10^b} in some order. It is also necessary (and
sufficient) that y/x lies in the interval [sqrt(3/2),sqrt(2)], or equivalently
that (y+x)/(y-x) lies in [3+sqrt(8),5+sqrt(24)] = [5.82842...,9.89897...].
Thus we need to make (5/2)^m*10^(a-b), or its reciprocal, in this range.
For each m there is clearly at most one power of 10 that will do. m=2,a=b
gives (105,145); m=3,b=a+2 gives (3375,4625), and so on.

There are infinitely many non-equivalent solutions, because log(5/2) / log(10)
is irrational.

One can use exactly the same argument to find squares whose initial 2 can 
be replaced by a 3, of course, except that the range of (y+x)/(y-x) changes.