A four-digit number has digits that follow each other in the number sequence. If we swap the digits in the hundreds and thousands places, we get a perfect square. Which number is this?
Let the sought four-digit number be

$$
1000 x+100(x+1)+10(x+2)+(x+3)
$$

After the rearrangement,

$$
1000(x+1)+100 x+10(x+2)+(x+3)=1111 x+1023=11(101 x+93)
$$

Among the prime factors of this number is 11, but then it can only be a perfect square if $11^2$ is also a divisor, meaning $101 x+93$ must be divisible by 11. However,

$$
\frac{101 x+93}{11}=9 x+8+\frac{2 x+5}{11}
$$

therefore, $2 x+5$ must be divisible by 11. Since $x$ is a single-digit number, then

$$
2 x+5=11
$$

from which

$$
x=3
$$

and thus the sought number is 3456. Indeed, $4356=66^2$.

Tamás Endrödy (Bp., III., Árpád g. I. o. t.)