Putnam 2008, Problem A-1

Let $f: \mathbb{R}^2 \to \mathbb{R}$ 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 $g: \mathbb{R} \to \mathbb{R}$ such that $f(x,y) = g(x) - g(y)$ for all real numbers $x$ and $y$.

Solution:
The function $g(x) = f(x,0)$ works. Substituting $(x,y,z) = (0,0,0)$ into the given functional equation yields $f(0,0) = 0$, whence substituting $(x,y,z) = (x,0,0)$ yields $f(x,0) + f(0,x) = 0$. Finally, substituting $(x,y,z) = (x,y,0)$ yields $f(x,y) = -f(y,0) - f(0,x) = g(x) - g(y)$.

**Remark:** A similar argument shows that the possible functions $g$ are precisely those of the form $f(x,0) + c$ for some $c$.