$S_n = x_1^3+x_2^3+ \cdots +x_n^3$ squares perfect

Question

It is considered arithmetic progression $x_1,x_2, \cdots, x_n,\cdots, x_1 \neq0$ . Show that if sums $$S_n = x_1^3+x_2^3+ \cdots +x_n^3$$ is squares perfect for any natural $n \in N$, then there are $k\in N^*$ so $x_n=nk^2$, for any $n \in N$. All my attempts were fruitless.

Answer 1

I'm not sure if this helps, but note that $$\sum_{m=1}^{n}{m^3}={(\frac{n(n+1)}2)^2}$$ Setting $x_i=nk^2$ we get $$S_n=\sum_{m=1}^{n}{(mk^2)^3}= k^6\sum_{m=1}^{n}{m^3}={(\frac{k^3n(n+1)}2)^2}$$ This is only the converse, however. I have no idea how to proceed for the forward direction.