"Determine all functions $\Bbb{Z}\to\Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$" I came across this problem:
Let $\Bbb{Z}$ be the set of integers. Determine all functions $f$:$\Bbb{Z}\to\Bbb{Z}$, such that for all integers $a$,$b \in \Bbb{Z}$
$f(2a)+2f(b)=f(f(a+b))$.
The unsatisfying solution that was presented was to substitute $a=0$ and $a=1$, and notice that $f$ is an arithmetic progression, and find both coefficients.
A satisfying solution, in my view, should go along these lines: What special property of $\Bbb{Z}\to\Bbb{Z}$ functions allows us to resolve equations involving such functions, and their convolutions?
For that I come for your help.
 A: Here is my approach (which comes after substituting all kinds of algebraic expressions for $a$ and $b$).
Let $P(a,b)$ denote the property "$f(2a)+2f(b)=f(f(a+b))$".
Then subtracting $P(a,0)$ from $P(0,a)$ gives \begin{equation}\label{1}\tag{1} f(2a)+f(0)=2f(a).\end{equation}
Let $z=f(0)$. Then $P(0,a)$ gives $f(0)+2f(a)=f(f(a))$, i.e. $z+2k=f(k)$ for all $k\in\operatorname{Im}(f)$. Inserting this into the original functional equation gives
\begin{equation}\tag{2}\label{2}
f(2a)+2f(b)=f(2a+2b+z)
\end{equation}
for all $a,b\in\mathbb Z$. From \eqref{1}, this is equivalent to $f(x)+f(y)+z=f(x+y+z)$, where $x=2a,y=2b$. This condition shall be called $Q(x,y)$. Now $Q(0,2n)$ and $Q(2,2n-2)$ for $n\in\mathbb Z$ give $f(2n)-f(2n-2)=f(2)-f(0)$ for all $n\in\mathbb Z$. It follows that \begin{equation}\label{3}\tag{3} f(2n)=z+n(f(2)-z).\end{equation}
From \eqref{1} with $a=n$ and $a=1$ we get $$f(n)=\frac{f(2n)+z}2=z+\frac n2(f(2)-z)=z+n(2f(1)-z).$$
Let $C=2f(1)-z\in\mathbb Z$. Putting that into $P(a,b)$ gives $$z+2aC+2z+2bC=z+C(z+C(a+b)),$$ i.e. $$2(a+b)C +2 z= C^2 (a+b)+Cz$$ for all $a,b\in\mathbb Z$. Let $a=-b$, then $$2z=Cz.$$
Therefore, $z=0$ or $C=2$. If $C=2$ then $z$ is arbitrary. If $z=0$ then the second-last equation for $a=1,b=0$ gives $2C=C^2$, i.e. $C=0$ or $C=2$. Therefore, the solutions to the original equation are only $$f(n)=z+2n,$$ where $z\in\mathbb Z$ is arbitrary OR $$f(n)=0$$ and it is easy to check that all such functions indeed solve the original equation.
