MHB Solve Functional Equation on $\mathbb{Z}$

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Let $ \mathbb{Z} $ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z} $ such that, for all integers $a$ and $b$, $f(2a)+2f(b)=f(f(a+b))$.

 

[anemone]

Spoiler: Solution of other:

Substituting $a=0$, $b=n+1$ gives $f(f(n+1))=f(0)+2f(n+1)$.

Substituting $a=1$, $b=n$ gives $f(f(n+1))=f(2)+2f(n)$.

In particular, $f(0)+2f(n+1)=f(2)+2f(n)$ and so $f(n+1)-f(n)=\dfrac{1}{2}\left(f(2)-f(0)\right)$.

Thus, $f(n+1)-f(n)$ must be constant. Since $f$ is defined only on $\mathbb{Z}$, this tells us that $f$ must be a linear function. Write $f(n)=Mn+K$ for arbitrary constants $M$ and $K$, and we need to only determine which choices of $M$ and $K$ work.

Now, $f(2a)+2f(b)=f(f(a+b))$ becomes $2Ma+K+2(Mb+K)=M(M(a+b)+K)+K$ which we may rearrange to form

$(M-2)(M(a+b)+K)=0$

Thus, either $M=2$ or $M(a+b)+K=0$ for all values of $a+b$. In particular, the only possible solutions are $f(n)=0$ and $f(n)=2n+K$ for any constant $K\in \mathbb{Z}$.