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 \(y\).

Proof:

We start with: \[f(x, y)+ f(y,z)+ f(z, x) = 0\]

Then we let \(x = y = z\), and we have: \[f(x,x) + f(x,x)+ f(x,x) = 3f(x,x) = 0\] \[f(x,x) = 0\]

Next, we let \(x = z\): \[f(x, y)+ f(y,x)+ f(x, x) = 0\] \[f(x, y)+ f(y,x) = 0\] \[f(x, y) = -f(y,x)\]

Rewriting our original equation: \[f(x, y) = -f(z,x) - f(y,z)\] \[f(x, y) = f(x,z) - f(y,z)\]

Finally, we simply define $g$: \[g(x) = f(x,z)\]