The Prisoner's Dilemma

Marion Cohen (Drexel University), Matthew Helmer (Pacific Lutheran University), and Robert Noll (PMP) solved this before we published a correction to an earlier
misprint (which substituted $f(x)-f(y)$ in place of $f(x-y)$, subtly changing the problem although not the answer). Nick Rauh (Seattle Universal Math Museum) solved it as stated here. Similar methods work for both versions, so we give just a solution for the one stated here.

This problem is an example of what’s called a “functional equation” problem: we have some general property satisfied by the function $f$ and want to conclude something about how to compute $f(x)$ itself. A common technique to use in such a case is to find substitutions for the variables in the functional equation that let you simplify the resulting expression to something useful. For example, if you try substituting $y=0$, you get $x(f(x)+f(0))=xf(x),$ which simplifies to $xf(0)=0$. Since this relationship has to hold for all $x$, we can conclude that $f(0)=0$, which is already something interesting.

About the next simplest value to try for $y$ is 1, which gives us that $(x-1)(f(x)+f(1))=(x+1)f(x-1)$. That doesn’t seem to give us a formula for any other values of f; it would be nice if there was a way to get rid of that leading factor of $x-1$. So we try a different substitution: $y=x-1$. (It might seem odd to just arbitrarily set $y$ to something in terms of $x$, but since the original functional equation is supposed to hold for all possible values of $x$ and $y$, it will certainly hold for $x=2$ and $y=1$ or $x=3$ and $y=2$, or really for any value of $x$ and $y=x-1$.) With this substitution, we get $f(x)+f(x-1)=(2x-1)f(1)$.

This equation still doesn’t look so promising for getting a formula for $f$, but if we step back and look at the last two equations we have obtained, we see that both of them involve just $f(x),f(x-1),$ and $f(1).$ Therefore, we can solve the first equation for $f(x-1)=(x-1)(f(x)+f(1))/(x+1)$ and substitute this value into the second equation. Simplifying the resulting equation gives us $xf(x)=x^{2}f(1)$. Whenever $x\ne 0$, we can cancel it to see that $f(x)=xf(1).$ But $f(1)$ is just some number, call it $k$. So we know that for all $x\ne 0$, $f(x)=kx$. But at the beginning we deduced that $f(0)=0=k0$ as well, so actually we know that for all $x$, $f(x)=kx$.

Conversely, if $k$ is an arbitrary real number, $(x-y)(kx+ky)=(x+y)k(x-y).$ Therefore, we conclude that the sign-flipping functions are exactly the functions of the form $f(x)=kx$ for some real number $k$.