(OIM 2005) Find all integers that are coprime with all terms of the sequence defined by $u_{n}=2^{n}+3^{n}+6^{n}-1$ for $n \in \mathbb{N}^{*}$.
The numbers 2, 3, 6 should remind you of something: they indeed satisfy $1 / 2 + 1 / 3 + 1 / 6 = 1$. This observation might seem useless since we are dealing with integers, and we cannot compute inverses (except for 1 and -1). However, recall that we also discussed inversion modulo a prime number. We are therefore working modulo $p$, where $p$ is a prime number. Fermat's Little Theorem implies that $a \times a^{p-2} \equiv 1 \pmod{p}$ for $a$ an integer coprime with $p$. Thus, $a^{p-2}$ plays the same role modulo $p$ as $1 / a$. This naturally leads us to consider $a_{p-2} = 2^{p-2} + 3^{p-2} + 6^{p-2} - 1$. However, be careful: if 2, 3, or 6 are not coprime with $p$, i.e., $p = 2$ or $p = 3$, we need to reason differently.

Let $p > 3$ be a prime number. We want to show that $p \mid a_{p-2}$. Similarly to working with fractions, we will try to compute $a_{p-2}$ modulo $p$ by "putting everything under the same denominator," or, equivalently, multiplying everything by $2 \cdot 3 \cdot 6$, since $2 \cdot 3 \cdot 6$ is not congruent to 0 modulo $p$. We then have $2 \cdot 3 \cdot 6 \cdot a_{p-2} \equiv 3 \cdot 6 + 2 \cdot 6 + 2 \cdot 3 - 2 \cdot 3 \cdot 6 \equiv 36 - 36 \equiv 0 \pmod{p}$ (this penultimate equality is not a coincidence; it is equivalent to saying that $1 / 2 + 1 / 3 + 1 / 6 = 1$). Thus, we have $p \mid a_{p-2}$, as desired.

But how will all of this help us with our problem? To simplify the notation, consider $A$ the set of integers coprime with all terms of the sequence. Let $m \in A$. As is often the case in arithmetic, we will try to reduce the problem to working with prime numbers, which have many useful properties (such as the one we just demonstrated). Here, it is very natural: the condition on $m$ means exactly that every prime factor $p$ of $m$ does not divide any of the terms of the sequence, i.e., $p \in A$. Since every prime number $q > 3$ satisfies $q \mid a_{q-2}$, $A$ contains no prime numbers, except possibly 2 and 3.

It remains to handle the cases of 2 and 3. We compute: $2^{1} + 3^{1} + 6^{1} - 1 = 10$, which is not coprime with 2, so 2 does not belong to $A$. Similarly, $2^{2} + 3^{2} + 6^{2} - 1$ is divisible by 3, so 3 does not belong to $A$. Finally, $A$ contains no prime numbers and thus $A = \{-1, 1\}$.

## 3 Group $C$: Inequalities and Functional Equations

## 1 Sunday 23 afternoon: Inequalities, Joon Kwon

We can refer to Guillaume Conchon-Kerjan's course from 2013.

## 2 Monday 24 morning: Gabriel Pallier

Warning This course revisits some elements from Pierre Bornsztein's course at the Saint-Malo 2003 stage, which readers can consult for a more developed treatment of the broad subject of solving functional equations.

Recall the definition of a real function of a real variable. $\mathbb{R}$ is the set of real numbers; intuitively, we identify it with the real line equipped with its origin.

Definition 152. Let $X$ be a non-empty subset of $\mathbb{R}$. Defining a function from $X$ to $\mathbb{R}$ means associating with every $x$ in $X$ a unique real number denoted $f(x)$. $X$ is called the domain of definition of $f$. We write this as

$$
f: X \rightarrow \mathbb{R}
$$

If $S$ is a subset of $X$, we call the restriction of $f$ to $S$ and denote it $f_{\mid S}$ the function that associates with every $s \in S$ the value $f(s)$.

We may replace the codomain $\mathbb{R}$ with one of its subsets $Y$. We then write $f: X \rightarrow Y$.

Definition 153. Let $f: X \rightarrow Y$. We say that $f$ is injective if

$$
\forall x, y \in X, \quad f(x) = f(y) \Longrightarrow x = y
$$

In other words, distinct numbers have distinct images.

Definition 154. Let $f: X \rightarrow Y$. We say that $f$ is surjective if

$$
\forall y \in Y, \quad \exists x \in X \mid f(x) = y
$$

In other words, all values in $Y$ are attained.

In a functional equation, the function is an unknown. It can be sought among the set of functions from $X$ to $\mathbb{R}$, which may satisfy certain properties (for example: among the functions that only take positive values). There can be multiple equations, or even multiple unknown functions.

Many common functions in mathematics (linear, affine, exponential, logarithmic, circular functions, factorial...) appear linked to a particular functional equation. We understand a function (or a family of functions) better the more we know about the functional equations that characterize it.

## The First Cauchy Equation

Here we seek to characterize all functions $f$ of the real variable such that

$$
f(x + y) = f(x) + f(y)
$$

for all $x$ and $y$ in a subset $X$ of $\mathbb{R}$. We do not yet fix $X$, but ideally, we want it to be as large as possible.

Remark 155. For $X$ to be such that the Cauchy equation (VII.1) makes sense, it must be that for all $x$ and $y$ in $X$, $x + y$ is also in $X$.