Solve $a b c-1=(a-1)(b-1)(c-1)$. With $a, b, c \geq 2$.
We are looking for $k a b c-1=k(a-1)(b-1)(c-1)$ such that we can notice that $a b c-1>a b c-b c>(a-1)(b-1)(c-1)$. Therefore, $k>1$. Idea: for $a, b, c$ large enough, then we have $a b c \sim(a-1)(b-1)(c-1)$, there is no solution. Let's show this rigorously. We can assume $a \leq b \leq c$

$$
\frac{a b c}{(a-1)(b-1)(c-1)}-\frac{1}{(a-1)(b-1)(c-1)}=k
$$

So in particular:

$$
1<k<\frac{a b c}{(a-1)(b-1)(c-1)}
$$

The function $\frac{x}{x-1}=1+\frac{1}{x-1}$ is decreasing, so we deduce that $\frac{a}{a-1} \leq \frac{b}{b-1} \leq \frac{c}{c-1}$. And thus

$$
1<k<\frac{a^{3}}{(a-1)^{3}}
$$

For $a \geq 5$, this gives $1<k<\frac{125}{64}$ and thus $1<k<2$ no solution. It is therefore sufficient to solve the cases $a=2,3,4$.

Case $a=4.1<k<\frac{64}{27} \leq 3$. So $k=2$. We are looking for $4 b c-1=6(b-1)(c-1)$. So $1=0[2]$ no solution.

Case $a=3.1<k<\frac{27}{8} \leq 4$ so $k=2$ or $k=3$.

For $k=3,3 b c-1=6(b-1)(c-1)$. So $-1=0[3]$ no solution.

For $k=2,3 b c-1=4(b-1)(c-1)$ and thus

$$
b c-4 b-4 c+5=0
$$

We can factorize

$$
(b-4)(c-4)=11
$$

Since 11 is prime, the solutions are thus $b=5$ and $c=15$.

Case $a=2,1<k<\frac{8}{1}=8.2 b c-1=k(b-1)(c-1)$. We deduce that $k$ is odd. For $k=3$. This gives

$$
b c-3 b-3 c+4=0
$$

So $(b-3)(c-3)=5$ with the solution $b=4$ and $c=8$.

For $k=5$. This gives

$$
3 b c-5 c-5 b+6=0
$$

So $9 b c-15 c-15 b+18=0$ thus $(3 b-5)(3 c-5)=7$ with the solution: $b=2$ and $c=4$.

Finally for $k=7$. We have

$$
5 b c-7 b-7 c+8=0
$$

And thus $25 b c-35 b-35 c+40=0$, so $(5 b-7)(5 c-7)=49-40=9$. We have $b=2$ and $c=2$. To conclude, the solutions are then $(2,2,2),(2,2,4),(2,4,8),(3,5,15)$ (and the permutations of these solutions)

## 2 Friday afternoon, 18th: Victor Vermès

This course recalls the notion of the order of an invertible modulo n, and Fermat's little theorem. The latter is generalized with the Euler's totient function, which is introduced and applied in an exercise session. The course and exercises correspond to the course given by Thomas Budzinski to group C at the Montpellier 2014 camp.

## 3 Saturday afternoon, 19th: Wassim Trabelsi

## Tool: the GCD

In a Diophantine equation, when we have a factorized expression, it is sometimes useful to find the GCD between the two numbers in question. We then obtain very useful information for solving the problem.

For example, if $a$ and $b$ are coprime and $a b=k^{2}$, then we can deduce that $a$ is a square and $b$ is a square.

If $\operatorname{GCD}(a, b)=2$ and $a b=k^{2}$, then $a$ is twice a square and $b$ is also twice a square.