What is the five-digit square number whose first digit is 2 and fourth digit is 5?
I. solution. It is obvious that the base $N$ of the square number $N^{2}$ is a three-digit number, so $N$ can be written in the form $N=A \cdot 10^{2}+B \cdot 10+C$, where $A, B, C$ are the digits of $N$. Thus, its square is

$$
N^{2}=A^{2} \cdot 10^{4}+2 A B \cdot 10^{3}+\left(2 A C+B^{2}\right) 10^{2}+2 B C \cdot 10+C^{2}
$$

Due to the second digit of $N^{2}$, $A^{2}<3$, so $A=1$. The fourth, fifth digit's place value is 10, for which only the $2 B C \cdot 10$ and $C^{2}$ terms have an influence. Among them, the first is divisible by 20, so the tens place of $C^{2}$ must be odd. Among the squares of the digits, only $C=4$ and $C=6$ are such. In both cases, the units digit is 6, so the last digit of $N^{2}$ is 6.

For $C=4$, the value of the last two terms is $80 B+16$ and this is of the form $100 k+56$, where $k$ is an integer. From this, $80 B=100 k+40$, simplifying to $8 B=10 k+4$, and since only $3 \cdot 8$ and $8 \cdot 8$ end in 4 in the multiplication table of 8, $B=3$ or 8. Thus, $N=134$ or 184. However, in their squares, the first digit is 1 or 3, so they do not match.

For $C=6$, similarly from $120 B+36=100 k+56$ we get $12 B=10 k+2$, so $B=1$ or 6, hence $N=116$ or 166. The first does not match because it is smaller than the above 134. 166 fits: $166^{2}=27556$, this is the sought number.

Imre Gáti (Pécs, Zipernovszky K. Mechanical Engineering High School II. grade)

Remark. Almost every solution arrived at the result in this way. Some elements of the following solution can only be found in the solution of László Pázmándi (Budapest, József A. High School II. grade).

II. solution. Based on the given digits, the sought square number $N^{2}$ is $20050 \leq N^{2}<30000$, so for its positive square root $142 \leq N \leq 173$.

The place value of the fourth digit is 10, so it belongs to the "two-digit ending" of $N^{2}$, or the number written with its last two digits, which is in a simple relationship with the two-digit ending of the base $N$ as follows. In the following, we will ignore numbers of the form $N=5 k$ because the ending of $N^{2}$ is 00 or 25, so its tens digit is not 5.

The two-digit ending of the squares of integers $a$ and $b$ is equal if and only if the difference $a^{2}-b^{2}$ ends in 00, i.e., $a^{2}-b^{2}=(a-b)(a+b)=100 k=2^{2} \cdot 5^{2} k$. Thus, both factors $a-b$ and $a+b$ must be even, because the sum and difference of two integers have the same parity - either both are even or both are odd - now, however, one of them must be even. The product's two 5 factors can only come from one of the factors, because the opposite assumption, that both factors are divisible by 5, leads to $a-b=10 m, a+b=10 n$ and consequently $a=5(m+n)$, which leads to the excluded case. Therefore, either $a-b$ or $a+b$ is divisible by 50. From $a-b=50 p$ we get $a=b+50 p$, or the two-digit endings of squares repeat every 50 on the integers, and from $a+b=50 r$ we get $(a+b) / 2=25 r$, or the two-digit endings of squares of numbers that are "mirrored" about a multiple of 25 are the same, i.e., if $a=25 r+s$, then $b=25 r-s$.

Accordingly, for example, from the squares of the integers from 1 to 24, not divisible by 5, we can get every two-digit ending different from 00 and 25. Among them, only one has a tens digit of 5, which is the two-digit ending $56$ from $16^{2}$. This two-digit ending is also given by every base ending in $16+50=66$, $50-16=34$, and $100-16=84$.

Accordingly, among the limits of $N$, 142 and 173, only $N=166$ can match, and since we have no further data (constraint) on it, it must match.