A six-digit square number, when divided into two-digit segments, reveals that the last two-digit number is the same as the middle two-digit number, while the first two-digit number completes the middle two-digit number to 100. Which is this square number?
I. solution: Denoting the sought square number by $n^{2}$, we can assume that $n$ is a positive integer, because a number and its negative have the same square. Since $n^{2}$ is a six-digit number, thus

$$
317 \leqq n \leqq 999
$$

Denoting the number formed by the last two digits of $n^{2}$ by $a$, according to the conditions of the problem,

$$
n^{2}=10000(100-a)+100 a+a=1000000-9899 a
$$

Here, the prime factorization of 9899 is $19 \cdot 521$, so expressing the last term from the derived equation,

$$
19 \cdot 521 a=1000000-n^{2}=(1000+n)(1000-n)
$$

We know that a product can only be divisible by a prime number if one of the factors is also divisible by that prime number. Applying this to 521 for the first factor,

$$
1317 \leqq 1000+n \leqq 1999
$$

it can only be divisible by 521 if it equals $3 \cdot 521=1563$; the second factor can only be divisible by 521 if it equals 521. In the first case,

$$
n=563 \quad n^{2}=316969
$$

satisfies the conditions of the problem, in the second case $a=479,1000+n=1479=3 \cdot 17 \cdot 29$ is not divisible by 19, so (1) cannot be satisfied. Therefore, the problem has one solution, as in (2).

Remark: Some started from the number $b$ formed by the first two digits of the square number. From the equation $n^{2}=9899 b+10100$ that can be derived, further conclusions can only be drawn if someone notices that 10100 can be expressed as the sum of a square number and a number divisible by 521:

$$
10100=16 \cdot 521+42^{2}
$$

and thus

$$
(n+42)(n-42)=521(19 b+16)
$$

This observation shows that sometimes even the fortunate choice of notation can be crucial in solving a problem.

Many started from what can occur as the last two digits of a square number, but did not reach a solution. This path is certainly more tedious than the above solution, but it is feasible, as the following solution shows:

II. solution: The last two digits of a number's square depend only on the last two digits of the number itself, in other words, if two numbers differ by a multiple of 100, then their squares also do, because

$$
(100 a+b)^{2}=10000 a^{2}+200 a b+b^{2}=100(100 a+2 a b)+b^{2}
$$

It is sufficient to examine the last two digits of numbers ranging from 0 to 100. But even the endings of numbers from 0 to 50 provide all possible endings, as those above 50 repeat in sequence the same endings, because

$$
(50+a)^{2}=2500+100 a+a^{2}=100(25+a)+a^{2}
$$

Moreover, the endings occur in pairs up to 50: the last two digits of the square of a number symmetric to 25 are the same. Indeed, a number $x$ and the number $50-x$ symmetric to 25 have the same last two digits in their squares, and

$$
(50-a)^{2}=2500-100 a+a^{2}=100(25-a)+a^{2}
$$

Thus, the possible endings (last two digits) of square numbers can be obtained by examining the last two digits of the squares of numbers from 0 to 25. This yields the following endings:

00010409162124252936414449566164697681848996.

For each of these, we can form the number described in the problem and decide by taking the square root whether it is a square number or not. However, we can somewhat simplify the calculation: the square of an even number must be divisible by the second power of 2, the square of a number divisible by 3 must be divisible by the second power of 3. It is known that the square of every odd number is of the form $8 k+1$. It is easy to see that the square of a number not divisible by 3 is always of the form $3 k+1$, so a square number cannot be of the form $3 k+2$, and similar constraints can be established with respect to other prime numbers.

The first and last of the listed endings fall out, because forming the number required by the problem's conditions does not result in a six-digit number. The numbers corresponding to 01, 09, 25, 41, 49, 81, and 89 give a remainder of 5 when divided by 8 and thus cannot be square numbers, while the numbers corresponding to 24 and 56 are not square numbers because they are divisible by 8 but not by 16. Considering divisibility by 3 and 9, the numbers corresponding to 04, 16, 61, and 76 give a remainder of 2 when divided by 3, and the numbers corresponding to 29 and 64 are divisible by 3 but not by 9, so they are not square numbers. Of the remaining numbers (those corresponding to 21, 36, and 44), taking the square root, we also find that 316969 is the only number that meets the conditions of the problem.