liekkas wrote: Find all functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, such that 1) $f(0,x)$ is non-increasing ; 2) for any $x,y \in \mathbb{R}$, $f(x,y)=f(y,x)$ ; 3) for any $x,y,z \in \mathbb{R}$, $(f(x,y)-f(y,z))(f(y,z)-f(z,x))(f(z,x)-f(x,y))=0$ ; 4) for any $x,y,a \in \mathbb{R}$, $f(x+a,y+a)=f(x,y)+a$ . Let $P_2(x,y)$ be the assertion $f(x,y)=f(y,x)$ Let $P_3(x,y,z)$ be the assertion $(f(x,y)-f(y,z))(f(y,z)-f(z,x))(f(z,x)-f(x,y))=0$ Let $P_4(x,y,z)$ be the assertion $f(x+z,y+z)=f(x,y)+z$ Let $h(x)=f(0,x)$, non increasing $P_4(x,y,-x)$ becomes $f(x,y)=h(y-x)+x$ $P_2(x,0)$ becomes assertion $R(x)$ : $h(-x)=h(x)-x$ $P_3(0,x,2x)$ becomes $(h(x)+x-h(2x))(h(2x)-h(x))=0$ and so : Since non increasing, we have $h(2x)\le h(x)<h(x)+x$ And so $h(2x)=h(x)$ $\forall x>0$ and so, since non increasing $h(x)=c$ constant $\forall x>0$ Let $x<0$, $R(x)$ becomes $c=h(x)-x$ and so $h(x)=x+c$ $\forall x<0$ Impossible since non increasing And so $\boxed{\text{No such function}}$

Sorry for a typo It should be $f(0,x)$ is non-decreasing. Fixed.

liekkas wrote: Find all functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, such that 1) $f(0,x)$ is non-decreasing ; 2) for any $x,y \in \mathbb{R}$, $f(x,y)=f(y,x)$ ; 3) for any $x,y,z \in \mathbb{R}$, $(f(x,y)-f(y,z))(f(y,z)-f(z,x))(f(z,x)-f(x,y))=0$ ; 4) for any $x,y,a \in \mathbb{R}$, $f(x+a,y+a)=f(x,y)+a$ . Let $P_2(x,y)$ be the assertion $f(x,y)=f(y,x)$ Let $P_3(x,y,z)$ be the assertion $(f(x,y)-f(y,z))(f(y,z)-f(z,x))(f(z,x)-f(x,y))=0$ Let $P_4(x,y,z)$ be the assertion $f(x+z,y+z)=f(x,y)+z$ Let $h(x)=f(0,x)$, non decreasing $P_4(x,y,-x)$ becomes $f(x,y)=h(y-x)+x$ $P_2(x,0)$ becomes assertion $R(x)$ : $h(-x)=h(x)-x$ Let $x\ne 0$ : $P_3(0,x,2x)$ becomes $(h(x)+x-h(2x))(h(2x)-h(x))=0$ and so : $\forall x\ne 0$, either $h(2x)=h(x)$, either $h(2x)=h(x)+x$ 1) If $\exists u>0$ such that $h(2u)=h(u)=a$ then $f(x,y)=a+\min(x,y)$

2) If $h(2x)\ne h(x)$ $\forall x>0$, then $f(x,y)=a+\max(x,y)$