Exercise 10. Let $k \geqslant 2$ be an integer. Find the smallest integer $n \geqslant k+1$ for which there exists a set $E$ of $n$ distinct real numbers, each of which can be written as the sum of $k$ other distinct elements of $E$.
Solution to Exercise 10 Let $n$ be an integer and $E$ a set as described in the statement. We sort the elements of $E$ in ascending order: these are $x_{1}, 2 x_{1}+2 x_{n}$. Consequently, $n \geqslant k+4$.

Conversely, if $n=k+4$, and if $E=\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ is a suitable set, it is necessary to identify, for any integer $i \leqslant n$, three distinct real numbers $x_{u}<x_{v}<x_{w}$ different from $i$ such that $x_{i}$ is the sum of the real numbers $x_{j}$ when $j \neq i, u, v, w$; or, alternatively, such that $x_{u}+x_{v}+x_{w}=s-2 x_{i}$. We then set $\ell=\lfloor k / 2\rfloor$, so that $\ell \geqslant 1$, and proceed by case analysis depending on the parity of $k$.

$\triangleright$ If $k$ is even, we set $E=\{-\ell-2,-\ell-1, \ldots,-2,-1,1,2, \ldots, \ell+1, \ell+2\}$. The sum of the elements of $E$ is $s=0$. We then choose the real numbers $x_{u}, x_{v}$, and $x_{w}$ as follows:

$\triangleright$ if $3 \leqslant x_{i} \leqslant \ell+2$, we set $x_{u}=-x_{i}, x_{v}=1-x_{i}$, and $x_{w}=-1$;

$\triangleright$ if $1 \leqslant x_{i} \leqslant 2$, we set $x_{u}=-3, x_{v}=-x_{i}$, and $x_{w}=3-x_{i}$;

$\triangleright$ if $-\ell-2 \leqslant x_{i} \leqslant-1$, we reuse the choices of $x_{u}, x_{v}$, and $x_{w}$ made for $-x_{i}$, then change their signs.

$\triangleright$ If $k$ is odd, we set $E=\{-\ell-2,-\ell-1, \ldots, \ell+1, \ell+2\}$; this is the same set as before, to which we have added the integer 0. The sum of the elements of $E$ is still $s=0$, and it suffices to handle the case where $x_{i}=0$, for which we choose $x_{u}=-2$, $x_{v}=-1$, and $x_{w}=3$.

In conclusion, the integer sought is $n=k+4$.

Grader's Comment The problem was rarely addressed. It's a pity, as it was possible to make many simple but valuable remarks. Some students correctly handled the case $k=2$, which is a good start.