Are there any digit-matching equalities similar to $(30+25)^{2}=3025$ among two-digit, or four-digit natural numbers?
I. solution: Denoting the two-digit numbers within the parentheses by $a, b$ (i.e., $0 \leq a, b \leq 99$), our requirement can be written as $(a+b)^{2}=100 a+b$, or equivalently, $(a+b)^{2}-(a+b)=99 a$. Using the notation $a+b=c$, this can be written as $c^{2}-c=c(c-1)=99 a$, where $c=a+b \leq 99$, because its square can be written with four digits. Thus, $c(c-1)$ must be divisible by 99, i.e., by $9=3^{2}$ and 11. Since $c$ and $c-1$ are relatively prime, only one of them is divisible by 3, and therefore one of them is divisible by 9.

If both 9 and 11 divide $c$, then $c=99$ and $a=98$, so $b=c-a=1=01$. Indeed, $(98+01)^{2}=99^{2}=9801$ shows the desired equality.

If 9 divides $c$ and 11 divides $c-1$: $c=9 x, c-1=11 y$, i.e., $9 x-1=11 y$, where $x, y$ are positive integers and $9 x<99$, $11 y<98$. It is easy to show that the only solution to these conditions is $c=45$, and in this case $a=45 \cdot 44 / 99=20, b=c-a=25$. Indeed, $(20+25)^{2}=45^{2}=2025$ also satisfies the condition.

If 9 divides $c-1$ and 11 divides $c$, then similarly $c=55, a=30$, and we arrive at the example shown. Since $c-1 \leq 98$, this factor cannot be divisible by both 9 and 11.

Remark. A similar but longer trial-and-error process leads to the following thought: $b$ can only be a two-digit number such that the tens and units digits of $b^{2}$ match those of $b$, i.e., $b^{2}-b=b(b-1)=100 x=2^{2} \cdot 5^{2} x$. From $b<100$, $b$ and $b-1$ can only be $0,25,50,75$. Direct trial-and-error shows that only $b=01$ and $b=25$ satisfy, and from the quadratic equation, we get $a=98$ and $a=30$ and 20.

Horváth Kálmán (Kaposvár, Táncsics M. g. I. o. t.)

II. solution: From the equation $(a+b)^{2}=100 a+b$, we have: $a=50-b \pm \sqrt{2500-99 b}$. The discriminant is positive only for $b \leq 25$, and from the requirement $2500-99 b=d^{2}$ (where $d$ is an integer), we have $2500-d^{2}=(50-d)(50+d)=99 b$, and thus $a=50-b \pm d$. Here, on the one hand, $d^{2}<2500, d<50$, and on the other hand, only one of $50-d$ and $50+d$ can be divisible by 3, because otherwise their sum: 100 would also be divisible by 3. From these, similar trial-and-error leads to the already seen representations.

Pallós Lajos (Pannonhalma, Bencés g. II. o. t.)

Remark. Many papers did not accept the representation $(98+01)^{2}=9801$ because the number 01 is "not two-digit." This is true, but the problem spoke of numbers that can be written with two digits, and single-digit numbers can also be written with two digits (numbering machines!).