IMO 2019 Shortlisted Problems

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 $f(n+1)-f(n)=\frac{1}{2}(f(2)-f(0))$. 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 only determine which choices of $M$ and $K$ work.

Now, (1) 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 $K\in \mathbb{Z}$, and these are easily seen to work.

---

Alternative solution.

Let $K=f(0)$. First, put $a=0$ in (1); this gives $$\begin{array}{c}f(f(b))=2f(b)+K\end{array}\text{\hspace{1em}(2)}$$ for all $b\in \mathbb{Z}$. Now put $b=0$ in (1); this gives $$f(2a)+2K=f(f(a))=2f(a)+K,$$ where the second equality follows from (2). Consequently, $$\begin{array}{c}f(2a)=2f(a)-K\end{array}\text{\hspace{1em}(3)}$$ for all $a\in \mathbb{Z}$. Substituting (2) and (3) into (1), we obtain $$\begin{array}{rl}f(2a)+2f(b)& =f(f(a+b))\\ 2f(a)-K+2f(b)& =2f(a+b)+K\\ f(a)+f(b)& =f(a+b)+K.\end{array}$$ Thus, if we set $g(n)=f(n)-K$ we see that $g$ satisfies the Cauchy equation $g(a+b)=g(a)+g(b)$. The solution to the Cauchy equation over $\mathbb{Z}$ is well-known; indeed, it may be proven by an easy induction that $g(n)=Mn$ for each $n\in \mathbb{Z}$, where $M=g(1)$ is a constant. Therefore, $f(n)=Mn+K$, and we may proceed as in Solution 1.