Which are the arithmetic sequences in which the sum of the first $n$ elements is a square number for every $n$?
Solution. Let the arithmetic sequence be $a_{1}=a, a_{2}=a+d, a_{3}=a+2 d, \ldots, a_{n}=a+(n-1) d$.

$$
S_{n}=\frac{a_{1}+a_{n}}{2} n=\frac{a+a+(n-1) d}{2} n=\left(a+\frac{(n-1) d}{2}\right) n, \quad(n=1,2, \ldots)
$$

According to the condition, $S_{1}=a=b^{2}$, where $b$ is a positive integer. Then

$$
S_{4}=\left(a+\frac{3 d}{2}\right) \cdot 4=4 a+6 d=2(2 a+3 d)
$$

Since $S_{4}$ is a perfect square, $d$ must be divisible by 2, that is, $d=2 e$, where $e$ is a positive integer.

Let $p$ be any prime number. Then

$$
S_{p}=p\left(a+\frac{(p-1) d}{2}\right)=p(a+(p-1) e)=p(a-e+p e)
$$

Since $\mathrm{S}_{p}$ is a perfect square, $a-e$ must be divisible by $p$. The only way $a-e$ can be divisible by every prime number is if $a-e=0$, that is, $a=e$. Therefore, $d=2 e=2 a$.

Thus, we have found that the first term of the sequence is $a=b^{2}$, and its difference is

$$
d=2 a=2 b^{2}
$$

Sequences with these properties satisfy the conditions of the problem.