. Find all sequences $\left(a_{n}\right)_{n \geqslant 1}$ of strictly positive real numbers such that for every integer $n$, we have:

$$
\sum_{i=1}^{n} a_{i}^{3}=\left(\sum_{i=1}^{n} a_{i}\right)^{2}
$$
. We will show by induction that $a_{n}=n$ for all $n$. For the initialization, we apply the condition of the statement with $n=1$ which gives $a_{1}^{3}=a_{1}^{2}$ and thus indeed $a_{1}=1$ since $a_{1}$ is strictly positive and therefore in particular non-zero.

Suppose now that $a_{i}=i$ for all $i<n$ and let's show that $a_{n}=n$. By hypothesis, we have:

$$
\left(1+2+\cdots+n-1+a_{n}\right)^{2}=1^{3}+2^{3}+\cdots+(n-1)^{3}+a_{n}^{3}
$$

which after simplification gives the equation $a_{n}^{2}=a_{n}+n(n-1)$. The latter can be factored as $\left(a_{n}-n\right)\left(a_{n}+n-1\right)=0$ and has the unique positive solution $a_{n}=n$, as desired.