Problem A1
Let f\colon R^2\rightarrow R be a function such that f(x, y) + f(y, z) + f(z, x) = 0 for all real numbers x, y, z. Prove that there exists a function g\colon R\rightarrow R such that f(x, y) = g(x) - g(y) for all real numbers x and y.
Solution
Setting x = y = z = 0 gives 3f(0, 0) = 0, making f(0, 0) = 0.
Let now y = z = 0. This leads to f(x, 0) + f(0, 0) + f(0, x) = 0, implying that f(x,0) = -f(0,x) for all real x.
Finally, let z = 0 to obtain f(x, y) + f(y, 0) + f(0, x) = 0. Which comes to f(x, y) = -f(y, 0) - f(0, x). Now setting g(x) = f(x,0) shows that f(x,y) = g(x) - g(y).
Remark
Problems like this where the unknown is a function are, naturally enough, called functional equations. These are seldom studied in high school, which is probably why many simple equations like this are offered at mathematical olympiads. Advanced Calculus branches into fields of Differential and Integral equations which all are specific functional equations.