(Short List IMO 2014)

Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that

$$
f(f(m)+n)+f(m)=f(n)+f(3 m)+2014
$$

for all integers $m, n$.

## - Solutions -
This exercise is the fourth algebra problem from the 2014 IMO shortlist. Let \( f(0) = c \) and \( g(n) = f(n) - f(0) \) for all \( n \). The functional equation can be rewritten as:

\[
g(g(m) + n + c) + g(m) = g(3m) + g(n) + 2014
\]

Setting \( m = 0 \) gives \( g(n + c) = g(n) + 2014 \), in particular, \( g(kc) = 2014k \) for all \( k \in \mathbb{Z} \). Taking \( n = -c \) gives \( g(g(m)) + g(m) = g(3m) \). Therefore, \( g(g(m) + n + c) = g(g(m)) + g(n) + 2014 \), and by replacing \( n \) with \( n - c \), we get \( g(g(m) + n) = g(g(m)) + g(n) \). Replacing \( m \) with \( c \), we obtain \( g(n + 2014) = g(n) + g(2014) \).

Thus, \( g(n + 2014c) = g(n) + c \cdot g(2014) \), but \( g(n + c) = g(n) + 2014 \), so \( g(n + 2014c) = g(n) + 2014^2 \). This implies that \( c \cdot g(2014) = 2014^2 \). Setting \( m = n = c \) in the initial equation gives \( g(2014 + 2c) = 4 \times 2014 \), so \( g(2014) = 2 \times 2014 \), thus \( c = 1007 \), and \( g(n + 1007) = g(n) + 2014 \) and \( g(1007k) = 2014k \) for all integers \( k \). For example, \( g(1007g(m)) = 2014g(m) \). On the other hand, the equation \( g(g(m) + n) = g(g(m)) + g(n) \) allows us to show \( g(kg(m)) = kg(g(m)) \) for all integers \( k \), so \( g(1007g(m)) = 1007g(g(m)) \). We deduce that \( g(g(m)) = 2g(m) \), and then \( g(3m) = 3g(m) \).

Now, we have \( 3^{\varphi(1007)} g(n) = g(3^{\varphi(1007)} n) \). And \( 3^{\varphi(1007)} = 1007\ell + 1 \) for some integer \( \ell \), by Euler's theorem (\( \varphi \) denotes the Euler's totient function), so

\[
3^{\varphi(1007)} g(n) = g(1007\ell n + n) = g(n) + 2014\ell n = g(n) + 2 \times (3^{\varphi(1007)} - 1) n
\]

Thus, \( g(n) = 2n \) for all integers \( n \).

The only solution is therefore \( f(n) = 2n + 1007 \), which is easily verified to satisfy the problem statement.

## 2 Arithmetic

## 1 Cyclotomic Polynomials

This course is mainly derived from the analogous course proposed during the summer olympic training organized in Montpellier in 2014 and its course notes. References concerning cyclotomic polynomials can be found on pages 116 to 141.

## 2 Various Exercises

This course consisted of a tutorial entirely made up of exercises drawn from shortlists of International Mathematical Olympiads. The goal was to see arithmetic exercises using small but well-chosen ideas: using moduli, comparisons, the pigeonhole principle, the Chinese remainder theorem, appropriate substitutions...

## - Exercise Statements -