Exercise 13. Determine all triplets of integers $(x, y, z)$ satisfying the following property:

$$
\operatorname{gcd}(x, y, z) < \operatorname{gcd}(x+y, y+z, z+x)
$$
Solution to Exercise 13 Note that since $\operatorname{pgcd}(x, y, z)$ divides each of $x, y, z$, it also divides $\operatorname{pgcd}(x+y, y+z, z+x)$.

We observe that if the triplet $(x, y, z)$ is a solution, then the triplets $(k x, k y, k z)$ are solutions for all $k \in \mathbb{N}^{*}$. Therefore, we can assume, by dividing each variable by $\operatorname{pgcd}(x, y, z)$, that the integers $x, y, z$ are coprime. In particular, they are not all even.

Let $d=\operatorname{pgcd}(x+y, y+z, z+x)$. Then $d$ divides $(x+y)+(x+z)-(y+z)=2 x$ and similarly $d$ divides $2 y$ and $2 z$. Therefore, $d$ divides $\operatorname{pgcd}(2 x, 2 y, 2 z)=2 \operatorname{pgcd}(x, y, z)=2$. We deduce that $d=1$ or $d=2$. But since $d>1, d=2$ so the integers $x, y, z$ are all of the same parity. Since they are not all even, they are all odd.

Conversely, if $x, y$ and $z$ are all odd, given that $\operatorname{pgcd}(x, y, z)$ and 2 are coprime and divide $\operatorname{pgcd}(x+y, y+z, z+x)$, we have

$$
\operatorname{pgcd}(x, y, z)<2 \operatorname{pgcd}(x, y, z) \leqslant \operatorname{pgcd}(x+y, y+z, z+x)
$$

The solution triplets are therefore the triplets $\left\{(k x, k y, k z), k \in \mathbb{N}^{*}, x, y, z\right.$ odd $\}$.

Graders' Comments

The graders were very satisfied with the different approaches of the students. A few errors are to be noted: it is not because $2 n \equiv k \mod a$ that $k$ is even and it is not because $\operatorname{pgcd}(x, y, z)=\operatorname{pgcd}(x+ y, y+z, z+x)$ that $(x, y, z)=(x+y, y+z, z+x)$.