Exercise 11. In a classroom, there are ten students. Aline writes ten consecutive integers on the board. Each student chooses one of the ten integers written on the board, such that any two students always choose two different integers. Each student then calculates the sum of the nine integers chosen by the other nine students. Each student whose result is a perfect square then receives a gift.

What is the maximum number of students who will receive a gift?

A perfect square is an integer of the form $n^{2}$, where $n$ is a natural number.
Solution to Exercise 11 The exercise asks to determine the maximum value of a certain quantity. The reasoning therefore consists of two steps, called analysis and synthesis.

Step 1, the analysis: We show that the number of students receiving a gift is necessarily less than or equal to four, regardless of the numbers chosen by Aline.

Let $n, n+1, \ldots, n+9$ be the ten numbers written on the board and let $S$ be their sum. If one of the integers is erased from the board, the sum of the remaining nine numbers is one of the integers:

$$
S-n, S-n-1, \ldots, S-n-9
$$

which are ten consecutive integers.

We then seek the maximum number of integers that can be perfect squares among ten consecutive integers.

First, since perfect squares are positive integers, a set of ten consecutive integers containing a negative integer contains fewer perfect squares than the set $\{0, \ldots, 9\}$. Next, note that among the integers from 0 to 9, there are four perfect squares, which are the integers $0,1,4$ and 9. Suppose there exist ten consecutive positive integers such that at least five of them are perfect squares. Then there exists a positive integer $a$ such that the numbers $a^{2},(a+1)^{2},(a+2)^{2},(a+3)^{2}$ and $(a+4)^{2}$ are among the ten consecutive integers. We then have $9 \geqslant(a+4)^{2}-a^{2}=8 a+16>9$, which is absurd.

Thus, the maximum number of integers that can be perfect squares among the ten integers $S-n, \ldots, S-n-9$ is 4, which means that at most four students will receive a gift.

Step 2, the synthesis: We show that the bound obtained in the analysis is achievable, that is, there exist ten consecutive integers for which the students can arrange to receive four gifts.

For this, we observe that the sum of nine of the ten integers $n, n+1, \ldots, n+9$ is always between

$$
n+(n+1)+\ldots+(n+8)=9 n+36 \quad \text { and } \quad(n+1)+(n+2)+\ldots+(n+9)=9 n+45
$$

Thus, the perfect squares chosen by the students are integers between $9 n+36$ and $9 n+45$. We saw during the analysis that the set $\{0, \ldots, 9\}$ contains 4 perfect squares. We therefore seek $n$ satisfying $9 n+36=0$ and $9 n+45=9$, which leads to $n=-4$.

Thus, if Aline writes the integers

$$
-4,-3,-2,-1,0,1,2,3,4,5
$$

the students can choose the integers $5,4,1$ or -4, so that the sum of the remaining nine integers will be respectively $0,1,4$ and 9.

The largest number of students who can receive a gift is four.

Comment from the graders: This exercise was approached by most students, who generally understood the substance of it, but it was surprisingly not well done. Most students (more than two-thirds) managed to find the configuration allowing 4 winning students - many students forgot to count 0 as a perfect square.

However, very few students managed to rigorously justify that the interval from 0 to 9 was the interval of 10 consecutive integers containing the most perfect squares. If the grading had taken more account of the ideas than the way they were written, we still expected more care in the justification of the arguments. Thus, the majority of the solutions were incomplete, even though they were awarded 7 points, and few students produced a truly rigorous proof of their assertion. It is a pity to see phrases such as "the squares get further and further apart," "the square function is increasingly increasing," "the squares are increasingly dispersed" replacing calculations, which are the only source of rigor here. We also note vocabulary problems, notably when students claimed that the square function was exponential, to cite just one example.