Saudi Arabia Mathematical Competitions

Without loss of generality, we may assume that $f(0)=0$. Let $y>0$ and $n$ a positive integer. For $x=ny$ we get $$2f(ny)=f((n+1)y)+f((n+2)y).$$ The sequence $a_n=f(ny),n=0,1,2,\dots$, satisfies the second order linear recursive relation $$a_{n+2}=-a_{n+1}+2a_n$$ with $a_0=f(0)=0,a_1=f(y)$. The characteristic equation is $t^2+t-2=0$, having the roots $t_1=1,t_2=-2$. It follows that $a_n=\alpha +\beta (-2{)}^n$, where $\alpha +\beta =0$ and $\alpha -2\beta =f(y)$. We get $$f(ny)=a_n=\frac{1-(-2{)}^n}{3}\cdot f(y)$$ In particular, we have $f(4y)=-5f(y)$. On the other hand, from (3), we obtain $f(2y)=-f(y)$, hence $f(4y)=-f(2y)=f(y)$, for any $y\ge 0$. That is $f(y)=0$ for any $y\ge 0$. Let $x\in \mathbb{R}$. Then $$2f(x)=f(x+|x|)+f(x+2|x|)=0$$ hence $f(x)=0$. The desired functions are all constant functions.

A simple checking shows that any constant function is solution to the functional equation.

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Alternative solution.

If $x_1,x_2\in \mathbb{R},x_1>x_2$, we can find $x\in \mathbb{R}$ and $y\ge 0$ such that $x_1=x+4y$ and $x_2=x+y$. Indeed, solving the system in $x$ and $y$, we get $$x=\frac{1}{3}(4x_2-x_1),y=\frac{1}{3}(x_1-x_2)$$ Replacing $y$ by $2y$ in the functional equation we obtain $$2f(x)=f(x+2y)+f(x+4y),$$ hence we get $$f(x+y)+f(x+2y)=f(x+2y)+f(x+4y)$$ that is $$f(x+y)=f(x+4y),x\in \mathbb{R},y\ge 0$$ From (1) it follows $f(x_1)=f(x_2)$, for any $x_1,x_2\in \mathbb{R}$, hence $f$ is constant function.

A simple checking shows that any constant function is solution to the functional equation.

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Alternative solution.

As in the previous solution, we have $$2f(x)=f(x+y)+f(x+2y)=f(x+2y)+f(x+4y),$$ hence, we get relation (1). Taking $x=-y,y\ge 0$, from (1) we obtain $f(0)=f(3y),y\ge 0$, that is $f$ is constant on $[0,\infty )$. Taking $x=-3y$, it follows $$f(-3y)=f(y)=0,y\ge 0$$ hence $f$ is constant on $(-\infty ,0]$. Therefore, all solutions are given by the constant functions.

A simple checking shows that any constant function is solution to the functional equation.