Exercise 13. Let $n$ be a strictly positive integer. Find the largest integer $k$, depending on $n$, such that there exists a subset $S$ of $\{1, \ldots, 2 n-1\}$ having $k$ elements and such that if $a, b$ and $c$ are three elements of $S$ for which $\mathrm{a}+\mathrm{b}=\mathrm{c}$, the equality $\mathrm{a}=\mathrm{b}$ is necessarily satisfied.
Solution to Exercise 13 The exercise asks us to find the largest integer satisfying a certain property, and it inevitably involves two parts: the analysis (if an integer $k$ satisfies the property, then $\mathrm{k} \leqslant \mathrm{c}$ with $\mathrm{c}$ the integer to be determined) and the construction (an explicit example of a set $A$ containing $c$ elements).

To guess the answer and to better understand the problem, we can start by examining the statement for small values of $n$.

For example, we can note that the set $A=\{1,3, \ldots, 2 \mathrm{n}-1\}$ of positive odd integers less than or equal to $2 n-1$ satisfies the property in the statement since the sum of two odd numbers is even. This set contains $n$ elements.

Similarly, the set $B=\{n, n+1, \ldots, 2 n-1\}$ of integers between $n$ and $2 n-1$ also satisfies the property since the sum of two elements is greater than or equal to $2 \mathrm{n}$. This set also contains $n$ elements.

We have constructed two sets with $n$ elements, and we can conjecture that $n$ is the largest integer satisfying the property.

Let $S$ be a subset of $\{1, \ldots, 2 \mathfrak{n}-1\}$ such that for any triplet of elements $(a, b, c)$ of $S$, if $a+b=c$ then $\mathrm{a}=\mathrm{b}$. Let $\mathrm{k}=|S|$ be its cardinality and $\mathrm{a}_{1}<\ldots<\mathrm{a}_{\mathrm{k}}$ its elements. By hypothesis, among the numbers $a_{k}-a_{i}$ with $1 \leqslant i \leqslant k-1$, at most one belongs to $S$. Indeed, the number $a_{k}-a_{i}$ belongs to the set $S$ only if $a_{k}=2 a_{i}$, and this equality is satisfied by at most one element of $S$. The set $S^{\prime}=\left\{a_{k}-a_{i}, 1 \leqslant i \leqslant k-1\right\}$ has size $k-1$, is included in $\{1, \ldots, 2 n-1\}$, and satisfies

$$
2 n-1 \geqslant\left|S \cup S^{\prime}\right|=|S|+\left|S^{\prime}\right|-\left|S \cap S^{\prime}\right| \geqslant k+k-1-1=2(k-1)
$$

We have obtained that $2(k-1) \leqslant 2 n-1$. By considering the parity of the two numbers, we even get that $2(k-1) \leqslant 2(n-1)$ and thus that $k \leqslant n$, as conjectured.

The largest integer sought is therefore the integer $\mathrm{n}$.

Comment from the graders: The exercise is well solved!