## Solution to Exercise 11

Note that $a^{2} \leqslant 108$ so $a \leqslant 10$. Furthermore, $97 \leqslant a^{2}+b^{2} \leqslant 2 a^{2}$ because Noémie has more sheep than Tristan, so $a^{2} \geqslant 48.5$, and since $a^{2}$ is an integer, $a^{2} \geqslant 49$ thus $a \geqslant 7$. Therefore, there are only a few cases to consider:

$\triangleright$ If $a=7, 97 \leqslant 49+b^{2} \leqslant 108$ so $48 \leqslant b^{2} \leqslant 59$. Since $b^{2} \leqslant a^{2}=49$, we can try to bound $b^{2}$ between two perfect squares: we have $3616=b^{2}$ and $a^{2}+b^{2}=97$ is indeed odd and between 97 and 108, so $(81,16)$ is a solution.

$\triangleright$ If $a=10, 97 \leqslant 100+b^{2} \leqslant 108$ so $b^{2} \leqslant 8$. Therefore, $b^{2}<9$, so $b<3$. In particular, $b=0,1$ or 2. If $b=0$ or $1, b^{2}<2$, so Tristan has strictly fewer than 2 sheep, which is impossible. If $b=2, a^{2}+b^{2}=104$ which is even, which is impossible, so there is no solution.

Thus, we obtain that Noémie has 81 sheep and Tristan has 16.
Exercise 11. Noémie and Tristan are shepherds. Each has a number of sheep that is a perfect square, that is, an integer that can be written in the form $n^{2}=n \times n$ with $n$ a positive or non-negative integer. We therefore denote $a^{2}$ as the number of sheep Noémie has and $b^{2}$ as the number of sheep Tristan has. After a quick count of the total number of sheep, they deduce that $97 \leqslant a^{2}+b^{2} \leqslant 108$. Determine the number of sheep each has, knowing that:

1. Noémie has strictly more sheep than Tristan;
2. each of them has at least 2 sheep;
3. the total number of sheep $a^{2}+b^{2}$ is odd.