# Arithmetic progression and square numbers

Let $(a_n)_{n \geq 1}$ be an arithmetic progression with all its terms in $\mathbb{N_{\geq1}}$. If $a_1^3+a_2^3+...+a_n^3$ is a square number for all $n \geq 1$, prove that there is a natural number $x \geq 1$ such that $a_n=nx^2, \: \forall n \geq 1$

Obviously, the first term must be a square number, so $a_1=x^2$. Let $d=a_{n+1}-a_{n}$. I tried replacing $a_2$ in $a_1^3+a_2^3$ in order to get some information about $d$ and maybe use some sort of induction after, but I failed