Let $\mathbb{Z}$ be the set of integers. Determine all functions $f:\mathbb{Z}\to \mathbb{Z}$ such that, for all integers $a$ and $b$,

Step 1 (Find f(0)): Setting $a=b=0$: $f(0)+2f(0)=f(f(0))$, so $f(f(0))=3f(0)$. Let $c=f(0)$.

Step 2 (Substitution $b=0$): $f(2a)+2c=f(f(a))$ for all $a$.

Step 3 (Substitution $a=0$): $c+2f(b)=f(f(b))$ for all $b$. Combining with Step 2 (replacing $a$ by $b$): $f(2b)+2c=c+2f(b)$, giving $f(2b)=2f(b)-c$.

Step 4 (Main substitution): From the original equation: $f(2a)+2f(b)=f(f(a+b))$. From Step 3: $f(f(a+b))=c+2f(a+b)$. So $f(2a)+2f(b)=c+2f(a+b)$. Using $f(2a)=2f(a)-c$: $2f(a)-c+2f(b)=c+2f(a+b)$, hence $f(a+b)=f(a)+f(b)-c$.

Step 5 (Cauchy equation): Define $g(x)=f(x)-c$. Then $g(a+b)=g(a)+g(b)$, so $g$ is additive over $\mathbb{Z}$. Thus $g(x)=kx$ for some integer $k$, giving $f(x)=kx+c$.

Step 6 (Back-substitute): Plugging $f(x)=kx+c$ into the original: $2ka+c+2(kb+c)=k(k(a+b)+c)+c$. This gives $2k(a+b)+3c=k^2(a+b)+kc+c$. Matching coefficients: $k^2=2k$ gives $k=0$ or $k=2$, and $kc+c=3c$ gives $kc=2c$, consistent with both.

Conclusion: $f(x)=0$ for all $x$ (when $k=0,c=0$), or $f(x)=2x+c$ for any integer constant $c$. $■$