660*. All digits of a certain four-digit number, which is a perfect square, can be decreased by the same natural number, not greater than the smallest of its digits, so that a four-digit number, also a perfect square, is obtained. Find all such numbers.
$\triangle$ Obviously, the bases of both squares are two-digit numbers.

Let the base of the second, smaller, square be denoted by $x$, and the first by $y$. Let $k$ be the natural number by which each digit of the first square is reduced. We get:

$$
\begin{aligned}
x^{2}=\overline{a b c d} & =1000 a+100 b+10 c+d \\
y^{2}=\overline{(a+k)(b+k)(c+k)(d+k)} & =1000(a+k)+100(b+k)+10(c+k)+(d+k)
\end{aligned}
$$

Subtract the first equation from the second:

$$
y^{2}-x^{2}=1000 k+100 k+10 k+k=1111 k
$$

Since $1111=11 \cdot 101$, the difference $y^{2}-x^{2}$ is divisible by 101. But 101 is a prime number, and the numbers $x$ and $y$ are two-digit, so $y-x<101$. Therefore, the sum $y+x$ is divisible by 101.

$$
(y+x): 101
$$

This is only possible when $y+x=101$. Hence, $y=101-x$. We will have:

$$
y^{2}-x^{2}=(101-x)^{2}-x^{2}=101^{2}-202 x
$$

Consider the equation

$$
101^{2}-202 x=1111 k
$$

and divide it by 101:

$$
101-2 x=11 k
$$

From the last equation, it is clear that $k$ is odd. Let's try all odd values of $k$ less than 9.

1) Let $k=1$. Then

$$
101-2 x=11, \quad x=45, \quad y=101-45=56
$$

2) Let $k=3$. In this case

$$
101-2 x=33, \quad x=34, \quad y=67
$$

3) Let $k=5$. Then

$$
101-2 x=55, \quad x=23
$$

But this value of $x$ is impossible, because

$$
23^{2}=529<1000
$$

4) Let $k=7$. This value of $k$ does not work because $x$ would be even smaller than in the third case.

Let's square the values of $y$ we found:

$$
56^{2}=3136, \quad 67^{2}=4489
$$

Answer: 3136, all digits need to be reduced by 1; 4489, all digits need to be reduced by 3.