Let's find a four-digit square number where two adjacent digits are equal!
I. solution: The base is a two-digit number and divisible by 11, because its square is divisible by it. $(10 a+a)^{2}=100 a^{2}+10 \cdot 2 a^{2}+a^{2}$. Since the last digit of $2 a^{2}$ is even, the last two digits of the square can only be even if the two digits of $a^{2}$ have the same parity. Such numbers are $a=2,8.22^{2}$ is still a three-digit number, 88 is the number we are looking for, $88^{2}=7744$.

Those who chose this path did not use all the possibilities and had to go through more complicated options.

Solved by: Bakonyi Kornélia, Bendzsák Z., Bodonyi J., Erdősy Gy., Frigyes Éva, Fülöp M., Gacsályi S. (junior), Gehér L., Hosszú M., Izsák I., Kovács J., László F., Lásztity R., Magyarósi B., Miskolczi Ida, Németh R., Turczi Gy., Ungár P., Vermes R.

II. solution: The base must be divisible by 11 and two-digit. $(11 a)^{2}=121 a^{2}$, where $a<10$. Let $a^{2}=$ $10 b+c, 121 a^{2}=100(12 b+c)+10(b+2 c)+c$. Here, $b+2 c$ must also end in $c$, so $b+c$ must be divisible by 10. This only follows for the single-digit numbers when $a=8$ and indeed $88^{2}=7744$.

Solved by: Silfen Péter.

Those who followed a similar path but in a more complicated way: Blaskó F., Gaál E., Korányi Á., Kővári T., Magyar Á. Sz., Reiner Éva, Sós Vera, Tóth K.

III. solution: The square number $a a b b=11(100 a+b)$ is divisible by 11, so the base must also be divisible by 11. But then the square must be divisible by $11^{2}$, so $100 a+b=99 a+(a+b)$ must also be divisible by 11, so $a+b$ must be. Since $a+b \leq 18$, $a+b$ can only be 11, $a a b b=11^{2}(9 a+1)$ can only be a perfect square if $9 a \mp 1=c^{2}$, $9 a=c^{2}-1=(c-1)(c+1)$. The difference between the two factors is 2, so only one can be divisible by 3, so it must also be divisible by 9. Since $c \leq 9$, then $c+1=9, c=8, a=7, b=4$. The number we are looking for is $88^{2}=7744$.

Réthy Eszter (Budapest, Veres Pálné Gymnasium IV. a. o.)

Bánó Klára (Budapest, Veres Pálné Gymnasium VI. o.)

Solved by: Szúcs L.

Solved similarly: Csík M., Csuhássy Edit, Földes P., Párkány M., Spitz Vera, Szabó A., Szentmártony Z., Szépfalussy P., Tamás I., Tarnóczi Z., Vata L. and Vékony Mária.

Many only submitted the final result. There were those who even revealed that they found the correct result by chance. These are not counted as solutions.