Find all prime numbers $p, q, r$ such that $p$ divides $1+q^{r}$, $q$ divides $1+r^{p}$, and $r$ divides $1+p^{q}$.

## 2 Solutions
We start by examining the condition "p divides $1+q^{r}$". It rewrites as $\mathrm{q}^{r} \equiv -1[\mathrm{p}]$ and thus implies, in particular, $\mathrm{q}^{2 r} \equiv 1[\mathrm{p}]$. Therefore, the order of $\mathrm{q}$ modulo $p$ is a divisor of $2 r$. Since $r$ is assumed to be prime, it is thus an element of the set $\{1,2, r, 2 r\}$. If we further assume that $p \neq 2$, we have $q^{r} \not \equiv 1[p]$, and thus the order of $q$ modulo $p$ is necessarily 2 or $2 r$. In the first case, $p$ divides $q^{2}-1=(q-1)(q+1)$, but cannot divide $q-1$ (since $q^{r} \equiv -1[p]$) and by Gauss's lemma we get $q \equiv -1[p]$. In the second case, by Fermat's theorem we get that $2 r$ divides $p-1$. By permuting the numbers $p, q$ and $r$, we of course get analogous consequences from the other two conditions " $\mathrm{q}$ divides $1+\mathrm{r}^{\mathrm{p}}$ ", and " $r$ divides $1+p^{q}$ ".

Now suppose that $p, q$ and $r$ are all odd, and to begin with that we are in the case where $q \equiv -1[p]$. The prime number $p$ cannot therefore divide $q-1$ (since it already divides $q+1$ and it is not 2). According to the results of the first paragraph, the condition " $\mathrm{q}$ divides $1+\mathrm{r}^{\mathrm{p}}$ " implies that $\mathrm{r} \equiv -1[\mathrm{q}]$. By applying the same argument again, we find that $p \equiv -1[r]$. However, the three previous congruences are not compatible. For example, they imply $q \geqslant p-1, r \geqslant q-1$ and $p \geqslant r-1$, but two of our primes cannot be equal, due to our condition " $q \equiv -1[p] $", and if we assume that $p$ is maximal, necessarily $q=p-1$, which is impossible since $p$ and $q$ are odd. We deduce that, always in the case where $p$, $q$ and $r$ are assumed to be odd, $2 r$ divides $p-1$. By cyclically permuting the variables, we similarly prove that $2 p$ divides $q-1$ and $2 q$ divides $r-1$. Thus $8 p q r$ divides $(p-1)(q-1)(r-1)$, which is clearly not possible given that $8 p q r>(p-1)(q-1)(r-1)$. Finally, there is no solution when $p, q$ and $r$ are all odd.

We now come to the case where one of these three numbers is equal to 2. By cyclically permuting $p, q$ and $r$ again, we can assume that it is $p$. The conditions of the statement then say that $q$ is odd, that $r^{2} \equiv -1[q]$ and that $2^{q} \equiv -1[r]$. By what was done in the first paragraph, this last congruence implies that $r=3$ or that $2 q$ divides $r-1$. The first case leads to $9 \equiv -1[q]$, which only occurs if $q=5$ since we have already ruled out the case $q=2$. We also verify that the triplet $(2,5,3)$ is indeed a solution. In the second case, now, the product $2 q$ divides $r-1$, but also $2\left(r^{2}+1\right)$ since we know that $r^{2} \equiv -1[q]$. Thus $2 q$ divides $2\left(r^{2}+1\right)-2(r+1)(r-1)=4$, which cannot happen.

In conclusion, there are exactly three solutions which are the triplets $(2,5,3),(5,3,2)$ and $(3,2,5)$.

## V. Third Period

Content of this section
1 Group A: Geometry ..... 222
1 Geometry Course: Angle Hunting, Power of a Point ..... 222
2 Geometry Exercise Session ..... 229
3 Geometry Course/Exercise Session: Transformations ..... 238
2 Group B: Geometry ..... 247
1 Geometry Course ..... 247
2 Geometry Exercise Session ..... 250
3 Geometry Course/Exercise Session ..... 256
3 Group C ..... 257
1 Arithmetic Course ..... 257
2 Arithmetic Exercise Session ..... 260
3 Arithmetic Course/Exercise Session ..... 263
4 Inequalities ..... 266
4 Group D ..... 272
1 Combinatorics Course: Finite and Infinite Graphs ..... 272
2 Combinatorics Exercise Session ..... 279
3 Combinatorics Course/Exercise Session ..... 285
4 Polynomials ..... 296

## 1 Group A: Geometry

## 1 Geometry Course: Angle Hunting, Power of a Point

- Angle Hunting -

It is often very practical in a geometry problem to know which angles are equal in the figure, or what are the relationships between different angles. Here are some techniques for hunting angles in a figure.

## Central Angle - Inscribed Angle

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-222.jpg?height=604&width=618&top_left_y=842&top_left_x=716)

Theorem 1.1. $\widehat{A C B}=\frac{1}{2} \widehat{A O B}$, the inscribed angle is half the central angle.

A very practical corollary is the following:

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-222.jpg?height=815&width=892&top_left_y=1731&top_left_x=538)

Corollary 1.2. All inscribed angles are equal, including the "limit" inscribed angle: the angle formed by the tangent.