We form a number from four consecutive digits, and by swapping the first two digits of this number, we get the square of a number. Which is this number?

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The sought number:

$$
1000(a+1)+100 a+10(a+2)+(a+3)=11(101 a+93)
$$

This expression can only be a perfect square if $101 a+93$ is divisible by 11. But

$$
101 a+93=99 a+88+2 a+5
$$

Since $a$ is a positive integer less than 10, $2 a+5$ is only divisible by 11 if $a=3$. Therefore, the sought number is

$$
4356=66^{2}
$$

(Denes König, Budapest.

Number of solutions: 46.