Challenging problem about $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f(x,y)+f(y,z)+f(z,x) = 0$ for all real $x, y, z$ Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that $f(x,y)+f(y,z)+f(z,x) = 0$ for all real numbers $x, y$, and $z$. Prove that there exists a function $g : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x,y) = g(x)−g(y)$ for all real numbers $x$ and $y$.
Attempt: Can we just set $z = 0$ and solve for $f(x,y)$?
 A: This will not be true in general but it will be true for this unusual function that has the unusual property that:
$f(x,y) + f(y,z) + f(z,x) = 0$.
This means that $f(x,y) = -f(z,x)- f(y,z) $.
This is true for all $z$ so if we could define $g(w) = -f(c,w)$ and $g(w) = f(w,z)$ for some constant $c$ we would be done. (As $g(x) - g(y) = -f(c,x) -f(y,c)= f(x,y)$.)
But that would require that there be some constant $c$ so that $-f(c,w) =f(w,c)$ for all $w$ and that is not true in general.
But, $f$ is not a usual function.  Maybe this IS true for $f$.
Can we prove that $f(w,z) = -f(z,w)$ for all $w,z$.?
Let $x = y = w$: Then $f(w,w) + f(w,z)+ f(z,w) = 0$ so $f(w,z) = -f(z,w) - f(w,w)$.
Shoot!  That was so close but $-f(z,w) - f(w,w) \ne -f(z,w)$ unless $f(w,w) = 0$ and that  is not true in general.
But, again, $f$ is not a general function.  Maybe $f(w,w)$ does equal $0$ for all $w$.
Let $x = y = z = w$:  Then $f(w,w) + f(w,w) + f(w,w) = 0$ so $f(w,w) = 0$.
That's it!  We are done.
.....
If we set $g(w) = f(w,c)$ for some constant $c$ we get:
$f(x,y) + f(y,c) + f(c,x) = 0$ so
$f(x,y) = -f(c,x) - f(y,c) = -f(c,x) - g(y)$.
Now $f(c,x)+ f(x,x) + f(x,c) = 0$ so
$f(c,x) = -f(x,c) - f(x,x) = -g(x) - f(x,x)$ so
$f(x,y) = -f(c,x) - g(y) = g(x) + f(x,x) - g(y)$.
Now $f(x,x) + f(x,x) + f(x,x) = 0$ so $f(x,x) = 0$ so 
$f(x,y) = g(x) +f(x,x) - g(y) = g(x) - 0 - g(y) = g(x) - g(y)$.
We are done.
Or for a third time:
$f(x,y) = -f(y,c) - f(c,x) $
$=-g(y) - (0 - f(x,x) - f(x,c))$
$= -g(y) + g(x) + f(x,x) $
$=g(x)-g(y) + \frac 13(f(x,x) + f(x,x) + f(x,x))$
$=g(x) - g(y) + \frac 13(0)$
$= g(x)-g(y)$. 
A: Hints:


*

*Prove that $f(x,x)=0$ for all $x$.

*Prove that $f(x,y)=-f(y,x)$ for all $x,y$.

*Set $g(x)=f(0,x)$, and then apply the initial attempt (since modified) to prove that $f(x,y)=g(x)-g(y)$.
