An exercice with MathJax
Problem 1: Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that
\begin{equation}
\forall (a, b) \in \mathbb{Z} \times \mathbb{Z}, \quad f(2a) + 2 f(b) = f(f(a + b))
\label{eq1}
\end{equation}
Substituting $0$ for $a$, $x$ for $b$ in $(\ref{eq1})$ yields:
\begin{equation}
\forall x \in \mathbb{Z}, \quad f(0) + 2 f(x) = f(f(x))
\label{eq2}
\end{equation}
Substituting $1$ for $a$, $x – 1$ for $b$ in $(\ref{eq1})$ yields:
\begin{equation}
\forall x \in \mathbb{Z}, \quad f(2) + 2 f(x – 1) = f(f(x))
\label{eq3}
\end{equation}
Combining $(\ref{eq2})$ and $(\ref{eq3})$ yields:
$$\forall x \in \mathbb{Z}, \quad f(x) = f(x – 1) + {{f(2) – f(0)} \over 2}$$
Thus $f$ is necessary of the form $f(x) = m x + n$ for some $(m, n) \in \mathbb{Z} \times \mathbb{Z}$.
Expanding $(\ref{eq1})$ yields:
$$\forall (a, b) \in \mathbb{Z} \times \mathbb{Z}, \quad m (2 – m)(a + b) + n(2 – m) = 0$$
It follows that necessarily either $m = 2$ or $m = n = 0$.
We can verify that these values actually give solutions to equation $(\ref{eq1})$.
Thus the set of functions $f$ solutions to problem 1 is exactly:
$$\{ f: x \mapsto 0 \} \cup \{f: x \mapsto 2x + n, n \in \mathbb{Z} \}$$
$\Box$
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