Let $f:\mathbb{R}^2\to\mathbb{R}$ such that $f(x,y)+f(y,z)+f(z,x)=0 \text{ } \forall \text{ } x,y,z \in \mathbb{R}.$ Show that there exists a $g:\mathbb{R}\to \mathbb{R}$ such that $f(x,y)=g(x)-g(y).$
This was question $A1$ on the $2008$ Putnam. My attempt is given below.
Put in $x=y=z$ to get $3f(x,x)=0.$ This means that $f(x,x)=0$ for all real $x.$ Now, put in $y=z$ to get $$f(x,y)+f(y,y)+f(y,x)=0=f(x,y)+f(y,x).$$ So, $f(x,y)=-f(y,x)$ for all real $x,y.$ In particular, $f(x,0)=-f(0,x).$ Now, let $g(x):=f(x,0).$ This definition makes sense as it's defined on $\mathbb{R}$ and gives out an output that's in $\mathbb{R}.$ Note that $f(x,y)+f(y,0)+f(0,x)=0.$ This gives $f(x,y)=-f(0,x)-f(y,0)=f(x,0)-f(y,0)=g(x)-g(y).$ But, $x$ and $y$ are arbitrary, so this identity holds for all real $x$ and $y,$ as desired.
Is this proof fine? I'm skeptical since my proof seems too simple to be the correct answer to a Putnam question. Another thing that troubles me is that we could've set $g(x):=f(x,a)$ for some real $a$ and gotten the same answer. Does this mean there are infinitely many such $g?$
Your solution is correct and there are infinitely many solutions. A useful resource for this and other Putnam problems is https://kskedlaya.org/putnam-archive/.
Looking at the equation $f(x, y) = g(x) - g(y)$, you can notice that if there is some solution for $g$ than $g + c$ (where $c$ is a constant) is also a solution, since the constants cancel each other out.
Noticing such things is often useful when solving functional equations and is a fairly common "trick". In general you could say something along the lines of: Since, if $h(x)$ is a solution, then $h(x) + c$ is also a solution, we can assume that WLOG $h(0) = 0$.