functional equations - Find all functions $f(f(f(...(f(x_1,x_2),x_3),...),x_{2016}))=x_1+x_2+...+x_{2016}$

I am trying to solve the functional equation:

Find all functions $f:\mathbb R^2\rightarrow \mathbb R$ such that for all $\left \{x_1,x_2,...,x_{2016} \right \}\subset \mathbb R$:
$$f(f(f(...(f(x_1,x_2),x_3),...),x_{2016}))=x_1+x_2+...+x_{2016}$$

My work so far:

$f(x,y)=f(0+0+..+0+x,y)$

$f(f(f(...(f(0,0),0),...),x),y)=0+0+...+0+x+y=x+y$

I need help here.

Answer

So lets assume $f$ is a real function an

$$f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016})= x_1+x_2+\ldots + x_{2016} \tag{1}$$

We now apply

$$f(.,x_{2017}) \tag{2}$$

on the LHS and RHS of $(1)$ and we get

$$f( f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016}) ,x_{2017}) = f(x_1+x_2+\ldots + x_{2016},x_{2017}) \tag{3}$$

If we change the variable names in $(1)$ , $x_{i}$ by $x_{i+1}$, we get

$$f(f(f(\ldots f(f(x_2,x_3),x_4)\ldots),x_{2017})= x_2+x_3+\ldots + x_{2017} \tag{4}$$

In $(4)$ now we substitute $x_2$ by $f(x_1,x_2)$ and get

$$f(f(f(\ldots f(f( f(x_1,x_2) ,x_3),x_4)\ldots),x_{2017})= f(x_1,x_2) +x_3+\ldots + x_{2017} \tag{5}$$

The LHS of $(3)$ and $(5)$ are the same. So the RHS of $(3)$ and $(5)$ must be equal and we get

$$f(x_1+x_2+\ldots + x_{2016},x_{2017}) = f(x_1,x_2) +x_3+\ldots + x_{2017} \tag{6}$$

Here we set $x_1=0$, $x_2=0$, ..., $x_{2015}=0$ and get

$$f(x_{2016},x_{2017})=f(0,0)+x_{2016}+x_{2017} \tag{7}$$

We can use $(7)$ to evaluate the nested function expression of the LHS of $(1)$. You can start with the outer or with the inner expression. Finally you will get

$$f( f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016}) ,x_{2017})= 2015 f(0,0)+x_1+x_2+x_3+\cdots+x_{2016} \tag{8}$$

The RHS of $(1)$ and $(8)$ are equal, so ist must be

$$2015f(0,0)=0 \tag{9}$$

From $(9)$ and $(7)$ we see that

$$f(x,y)=x+y \tag{10}$$

as expected.