Putnam 2008

Yesterday was the first Saturday of December, hence the day of the Putnam National Mathematics exam. We had nine students take the exam, but we won't know results until (probably) late March.

For those who don't have immediate access to the exam, here is one of the problems:

Show that if f:R² &rarr R is a function that satisfies the relationship

f(x,y) + f(y,z) + f(z,x) = 0

then f(x,y) = g(x) - g(y) for some function g:R &rarr R.