(Olympiad Question) $f(2x) + 2f(y) = f(f(x+y))$ This was Problem 1 from the 2019 International Mathematics Olympiad.

Find all functions $f$ : $\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$  that satisfy $f(2x) + 2f(y) = f(f(x+y))$ whenever $x , y$ $\in$ $\mathbb{Z}$.

And this is my progress...

Let $x=y$
then $f(2x) + 2f(y) = f(f(x+y))$
then $f(2x) + 2f(x) = f(f(2x))$ (substitute $x$ with $y$)
then $2f(x) + 2f(x) = f(2f(x))$ (from Cauchy's theory $f(nx) = nf(x)$ where $n > 0$)
then $4f(x) = 2f(f(x))$ (from Cauchy's theory $f(nx) =nf(x)$ where $n > 0$)
then $2f(x) = f(f(x))$ (divide both sides by $2$)
then $f(x) = 2x$ which is one of the satisfied fuctions.
Let $x = -y$
then $f(2x) + 2f(y) = f(f(x+y))$
then $f(-2y) + 2f(y) = f(f(-y+y))$ (substitute $x$ with $-y$)
then $2f(-y) + 2f(y) = f(f(-y+y))$ (from Cauchy's theory $f(nx)=nf(x)$ where $n>0$ )
then $2f(-y) + 2f(y) = f(f(0))$ (from additional inverse property of any integers)
then $2f(-y) + 2f(y) = 0$ (from Cauchy's theory $f(0)=0$ )
then $f(-y) = -f(y)$ (from additional inverse property of any integers, since $f(y)$ $\in$ $\mathbb{Z}$)
then all functions $f$ : $\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$ that satisfy $f(2x) + 2f(y) = f(f(x+y))$ whereas $x,y$ $\in$ $\mathbb{Z}$ are odd.

I can't continue because I'd almost lost my dinner when I solve this, so I decided to take some Clorazepate and sleep.
Can you continue or check my progress? Is it right or not? Can you help me please?
 A: Another solution: 
Note that substituting $a=0$ implies that $$f(f(b))=f(0)+2f(b).$$ So setting $b=0$ implies $f(2a)+2f(0)=f(f(a))=f(0)+2f(a) \implies f(2a)=2f(a)-f(0).$
Then rewriting the original equation with these two properties in mind, $$\begin{align*} f(2a)+2f(b)&=f(f(a+b)) \\\iff 2f(a)-f(0)+2f(b)&=f(0)+2f(a+b) \\ \iff 2f(a)+2f(b)  &=2f(a+b)+2f(0) \\ \iff f(a)+f(b)&=f(a+b)+f(0)\end{align*}$$ Now define a new function $g(x)=f(x)-f(0)$ to see that $g(a)+g(b)=g(a+b)$, and hence $g$ satisfies Cauchy's functional equation. Thus $g(x)=mx$ for some constant $m$, and so $f(x)=mx+n$ for a constant $n=f(0)$. 
Putting this into the original equation yields that the only solutions are $f(x)=0$ and $f(x)=2x+n$ for an integral constant $n$, and we have already verified that these two work. 
A: For an alternative approach, notice that
$$f(2(x-1))+2f(y+1)=f(f(x+y))=f(2x)+2f(y)$$
so that
$$2(f(y+1)-f(y))=f(2x)-f(2x-2).$$
Setting $x=0$ say gives $f(y+1)-f(y)=C$ is a constant.
Therefore $f(x)=Cx+D$ for constants $C$ and $D$.
Now put this into the original equation to find all possible
pairs $(C,D)$, etc.
