Putnam 1971 
Topic: Algebra Analysis

Problem: Let $F(x)$ be a real valued function defined for all real $x$ except for $x=0$ and $x=1$ and satisfying the functional equation $F(x) + F\left(\frac{x-1}{x}\right) = 1 + x$. Find all functions $F(x)$ satisfying these conditions.

Solution: In the given functional equation \[(1)\ F(x) + F\left(\frac{x-1}{x}\right) = 1 + x\] we substitute $\frac{x-1}{x}$ for $x$, obtaining \[(2)\ F\left(\frac{x-1}{x}\right) + F\left(\frac{-1}{x-1}\right) = \frac{2x-1}{x}.\] Also in (1), we substitute $\frac{-1}{x-1}$ for $x$ and obtain \[(3)\ F\left(\frac{-1}{x-1}\right) + F(x) = \frac{x-2}{x-1}.\] Adding (1) and (3) and subtracting (2) gives \[2F(x) = 1 + x + \frac{x-2}{x-1} - \frac{2x-1}{x} = \frac{x^3 - x^2 - 1}{x(x-1)}.\] \[F(x) = \boxed{\frac{x^3 - x^2 - 1}{2x(x-1)}}.\] That $F(x)$, defined in (4), does satisfy the given functional equation is easily verified. Therefore (4) is the only solution of the problem.