*Problem A1*

*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.