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.

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