Exercise 1. Find all quadruplets ( $p, q, r, \mathfrak{n}$ ) of strictly positive integers satisfying the following three conditions:

- $p$ and $q$ are prime,
- $p+q$ is not divisible by 3,

$-p+q=r(p-q)^{n}$.
Solution to Exercise 1: We first assume that $p \neq 3$ and $q \neq 3$, and consider $p$ and $q$ modulo 3. Since $p+q \not \equiv 0[3]$, we do not have $p \equiv 1[3]$ and $q \equiv 2[3]$, nor $p \equiv 2[3]$ and $q \equiv 1[3]$. Therefore, we are assured that $p \equiv q \equiv 1[3]$ or $p \equiv q \equiv 2[3]$. In both cases, 3 divides $p-q$ and thus divides $p+q = r(p-q)^n$ since $n \geqslant 1$. This is absurd and invalidates our assumption, showing that $p=3$ or $q=3$.

On the other hand, since $n \geqslant 0$, we know that $p-q$ divides $p+q$, so $p \neq q$, which means that $p$ and $q$ are coprime. Since $p-q$ divides $(p+q)+(p-q)=2p$ and $(p+q)-(p-q)=2q$, we know that $p-q$ divides $\operatorname{GCD}(2p, 2q) = 2 \operatorname{GCD}(p, q) = 2$. It follows that $p-q \in \{-2, -1, 1, 2\}$. Therefore, $(p, q) \in \{(2,3), (3,2), (3,5), (5,3)\}$; we treat these four cases one by one:

— if $(p, q) = (2,3)$, the equation becomes $5 = (-1)^n r$; its solutions are $r=5$ and $n \in 2\mathbb{N}^{*}$;

- if $(p, q) = (3,2)$, the equation becomes $5 = 1^n r$; its solutions are $r=5$ and $n \in \mathbb{N}^{*}$;

— if $(p, q) = (3,5)$, the equation becomes $8 = (-2)^n r$; its solutions are $(r, n) = (2,2)$;

- if $(p, q) = (5,3)$, the equation becomes $8 = 2^n r$; its solutions are $(r, n) = (1,3), (2,2)$, and $(4,1)$.

In conclusion, the solutions are: $(2,3,5,2n)_{n \in \mathbb{N}^{*}}, (3,2,5, n)_{n \in \mathbb{N}^{*}}, (3,5,2,2), (5,3,1,3), (5,3,2,2)$, and $(5,3,4,1)$.