Olympic Problem on Functional Equations

$f(2019) = f(f(0)) = f(f(0+0)) = f(2\times0) + 2f(0) = 3f(0) = 3\times2019 =\boxed{6057}$

Replacing in the given functional equation the $a$ by 0 and the $b$ by $a+b,$ we obtain that $f(0)+2f(a+b)=f(f(a+b)).$ This equation and the one given in the problem imply that $f(2a)+2f(b)=f(0)+2f(a+b)$ for any integer numbers $a$ and $b.$ Now, making $a=1,$ we get $f(2)+2f(b)=f(0)+2f(b+1)$ for all integers b. From this equation, we obtain that $f(b+1)-f(b)=\frac{f(2)-f(0)}{2}.$ Thus it can be proved by mathematical induction that $f(n)=c n+r,$ where $c$ and $r$ are constant integers. Substituting this expression into the given functional equation, it is obtained that $2c(a+b)+3r=c^2(a+b)+cr+r.\:\:\:\:\:\:(*)$ Making $a+b=0,$ moving all the terms to the left side and factoring, it follows the equation $(2-c)r=0.\:\:\:\:\:(**)$ Then there are two cases: $r=0$ or $r\neq0.$

Case 1: When $r=0.$

Substituting into equation (*), assuming that $a+b\neq 0,$ we get $2c=c^2.$ Then $c=0$ or $c=2.$ Both values work, so we obtain that two of the possible solutions of the given equation are $f(n)=0,$ or $f(n)=2n,$ for all integer values of $n.$

Case 2: When $r\neq0.$

The equation $(**)$ implies that $c=2.$ Then it is easy to verify that making $c=2$ and for any integer $r$ the equation $(*)$ is true. Therefore, the remaining solutions of the given equation will be any function of the form $f(n)=2n+r,$ where $r$ is any integer different from zero. We conclude that any solution of the given functional equation will be of the form $f(n)=0$ for any integer $n,$ or the form $f(n)=2n+r,$ where $n$ is any integer and $r$ is an arbitrary integer constant. Now, from the condition that $f(0)=2019,$ it follows that the corresponding solution has the form $f(n)=2n+2019,$ for any integer $n.$ Then $f(2019)=2(2019)+2019=\boxed{6057.}$

Note: This problem was inspired by the first problem of the IMO2019.