Let the six-digit square number be denoted by $n^{2}$, the two-digit number forming the middle part of this by $x$, and the difference between the last and the middle two-digit number by $y^{2}$. Then the requirement of the problem can be written as:

$$
n^{2}=x \cdot 10^{4}+x \cdot 10^{2}+\left(x+y^{2}\right)=10101 x+y^{2}
$$

By rearranging and factoring the coefficients, this can be brought to the following form:

$$
(n+y)(n-y)=37 \cdot 13 \cdot 7 \cdot 3 \cdot x
$$

$n^{2}$ is a six-digit number, which means that $10^{5} \leqq n^{2}<10^{6}$. Since $316^{2}<10^{5}<317^{2}$, it follows that

$$
317 \leqq n \leqq 999
$$

Furthermore, $x$ and $x+y^{2}$ are two-digit numbers:

$$
x \geqq 10 \text { and } x+y^{2} \leq 99
$$

thus

$$
y^{2} \leqq 99-x \leqq 99-10=89
$$

and since it is sufficient to consider non-negative values of $y$, we have $0 \leqq y \leqq 9$. Therefore, the factors on the left side of (7) are subject to the following constraints:

$$
\left\{\begin{array}{l}
n+y \leqq 999+9=1008 \\
n-y \geqq 317-9=308
\end{array}\right.
$$

and for the difference of the two factors:

$$
(n+y)-(n-y)=2 y \leqq 18
$$

Accordingly, we need to find a two-digit number $x$ such that the right side of (7) can be factored into two integer factors that satisfy the constraints (9) and (10), and from which $n$ and $y$ are integers, and finally, (8) is also satisfied for $x$ and $y$. For this purpose, we try to factor the right side of (7) in every possible way into two factors, also considering the factorization of $x$, let $x=x_{1} x_{2}$ (one of $x_{1}$ and $x_{2}$ can be 1).

The product of the prime factors under (7) is greater than 1008, so they cannot all be in the same factor. The product can be factored into two factors less than 1008 in the following ways:

$$
13 \cdot 777, \quad 21 \cdot 481, \quad 37 \cdot 273, \quad 39 \cdot 259 \quad \text { and } \quad 91 \cdot 111 .
$$

Looking for the two factors on the left side of (7) in the form $13 x_{1}$ and $777 x_{2}$, only $x_{2}=1$ satisfies (9), this factor is 777, so according to (10) the other factor must satisfy

$$
777-18=759 \leqq 13 x_{1} \leqq 795=777+18
$$

dividing by 13 and writing only the integer part of the quotient,

$$
58<x_{1} \leqq 61
$$

$n$ and $y$ are only integers if their sum and difference have the same parity. The factor 777 is odd, so only $x_{1}^{\prime}=59$ and $x_{1}^{\prime \prime}=61$ are possible.

In the first case, $x=x^{\prime} x_{2}=59$, the other factor is $13 \cdot 59=767$, smaller than 777, so $n+y=777, n-y=767$; from which $n=772, y=5$; and $x+y^{2}=84$, a two-digit number, so we have found a solution. Indeed, $772^{2}=595984$, which satisfies the condition.

In the second case, $x=61$, and $y=8$ is obtained, from which $x+y^{2}$ is a three-digit number, so no solution is obtained.

With similar reasoning, looking for factors in the form $21 x_{1}, 481 x_{2}$, only $x_{2}=1$ or 2 are possible according to (9). A solution is obtained in both cases: for $x_{2}=1$, $x_{1}=23, x=23, n=482, y=1$, and $n^{2}=232324$; and for $x_{2}=2$, $x_{1}=46, x=92, n=964, y=2$, and $n^{2}=929296$. (In the latter, $x_{2}, x_{1}, n$ and $y$ are twice, and $n^{2}$ and $x$ are four times the corresponding numbers of the previous solution.)

If we look for the factors in the form $37 x_{1}$ and $273 x_{2}$, only $x_{2}=2$ and 3 are possible according to (9). For $x_{2}=2$, $x_{1}$ can only be even, but $546-18=528$ and $546+18=564$ do not include an odd multiple of 37: $555=15 \cdot 37$. No solution is obtained for $x_{2}=3$ either, because in this case $x_{1}$ is odd, but $3 \cdot 273-18=801$ and $3 \cdot 273+18=837$ do not include an even multiple of 37: $814=22 \cdot 37$.

No solution is obtained with factors in the form $39 x_{1}, 259 x_{2}$, nor with factors in the form $91 x_{1}, 111 x_{2}$, because taking the values of $x_{2}$ allowed by (9) and the limits obtained from (10), either no multiple of the coefficient of $x_{1}$ falls within these limits, or the parity of the multiple obtained does not match the parity of $x_{2}$.

Accordingly, the following three square numbers satisfy the conditions of the problem:

$$
482^{2}=232324, \quad 772^{2}=595984 \quad \text { and } 964^{2}=929296
$$
A six-digit square number is cut into three two-digit numbers. Among these three two-digit numbers, the middle one is equal to the one before it, and the difference between the last and the middle two-digit number is a square number. Which is the original six-digit number?