Baltic Way 2019

Let us define an auxiliary function $g:\mathbb{R}\to \mathbb{R}$ by setting $$g(x)=f(f(x))+4f(x)+4x$$ for all real numbers $x$. The given functional equation may be written as $$2g(x)=-g(-x),$$ for all real numbers $x$, so that $$4g(x)=-2g(-x)=-(-g(-(-x)))=g(x),$$ for all real numbers $x$. We obtain from this that $$f(f(x))+4f(x)+4x=g(x)=0,$$ for all real numbers $x$. Let now $x$ be an arbitrary nonzero real number, and let us define a sequence $a_0,a_1,a_2,\dots$ by setting $$a_0=x,a_1=f(x),a_2=f(f(x)),a_3=f(f(f(x))),\dots$$ so that $$a_{n+1}=f(a_n)=f(f(a_{n-1}))=-4f(a_{n-1})-4a_{n-1}=-4a_n-4a_{n-1},$$ for each positive integer $n$. The general term of the sequence may be written as a function of $n$ through $$a_n=(A+Bn)(-2{)}^n,$$ which holds for all nonnegative integers $n$, and where $$A=a_0=x\text{and}B=-\frac{a_1}{2}-a_0=-\frac{f(x)}{2}-x.$$ If $B\ne 0$, then we may choose a positive integer $n$ so that $|A+Bn|>1$ and $2^n>2019$, and thus $$|f(a_{n-1})|=|a_n|=|A+Bn|\cdot |(-2{)}^n|>2019,$$ against the fact that $f(a_{n-1})\in [-2019,2019]$. Therefore we must have $B=0$. But now we may choose a positive integer $n$ so that $|x|2^n>2019$, leading to $$|f(a_{n-1})|=|a_n|=|a_0|\cdot |(-2{)}^n|=|x|2^n>2019,$$ which again produces a contradiction. We conclude that no function with the desired properties exists.

If $f$ satisfies the problem condition, then for all $x\in \mathbb{R}$ we have $$\begin{array}{rl}4|x|& =|2f(f(x))+f(f(-x))+8f(x)+4f(-x)|\\ & \le 2|f(f(x))|+|f(f(-x))|+8|f(x)|+4|f(-x)|\\ & \le 2\cdot 2019+2019+8\cdot 2019+4\cdot 2019=30285.\end{array}$$ This obviously does not hold for the big enough value of $x$.