Putnam 2009 A 1

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.