A four-digit square number has the sum of its two middle digits equal to its first digit from the left, and the sum of its first and third digits is ten times its last digit. Which is this square number?
Let the sought square number be

$$
1000 a+100 b+10 c+d
$$

where $a, b, c, d$ are natural one-digit numbers, of which $b, c, d$ can be zero.

According to the problem,

$$
b+c=a
$$

and

$$
a+c=10 d \text {. }
$$

Since $a+c \leq 18$, and $a \neq 0$, therefore

$$
\begin{gathered}
d=1 \\
a+c=10
\end{gathered}
$$

and from (1) and (4)

$$
2 a=10+b
$$

Let the square root of the sought square number be $10 x+y$, then $y$ can only be 1 or 9 due to (3), which implies that $1000 a+100 b+10 c$ is divisible by 4, which is only possible if $c$ is even. But if $c$ is even, then according to (4) $a$ is also even and thus according to (1) $b$ is also even.

From (5) $a$ can only be 6 or 8 and accordingly $b$ can only be 2 or 6 and thus according to (4) $c(=10-a)$ can only be 4 or 2. Therefore, only 6241 and 8621 can be considered, but among these only

$$
6241\left(=79^{2}\right)
$$

is a square number.