International Mathematical Olympiad (IMO) $2019$ , Day $1$ , Problem $1$ of $6$

Solution $1$ : Solution $2$ :

Let us examine two Cases:

CASE I ( $f$ is constant): This yields $C + 2C = C \Rightarrow 2C = 0 \Rightarrow C = 0$ . Hence, $\boxed{f(n) = 0}$ is the only constant solution for all $n \in \mathbb{Z}.$

CASE II ( $f$ is non-constant): The function $f(n)$ is linear in order to preserve the mapping of this Cauchy functional equation (i.e. $f(n) =An+B$ for $A,B \in \mathbb{Z}.$ ) Substituting this linear expression into the original equation above yields:

$[A(2a) + B] + 2[A(b)+B] = A[A(a+b)+B] + B;$

or $2Aa + 3B + 2Ab = A^{2}a + A^{2}b + (A+1)B.$

If $2A = A^2,$ then $A = 0,2$ . For $A = 0 \Rightarrow 3B = B \Rightarrow B = 0$ , which corroborates with Case I above. For $A = 2 \Rightarrow 4a + 3B + 4b = 4a + 4b + 3B \Rightarrow B = K$ , where $K$ is any integer. Thus, $\boxed{f(n) = 2n + K}$ is the family of solutions that satisfies non-constant $f$ .