2015 USAMO Problems/Problem 4

Problem

Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ such that $f(x)+f(t)=f(y)+f(z)$ for all rational numbers $x<y<z<t$ that form an arithmetic progression. ( $\mathbb{Q}$ is the set of all rational numbers.)

Solution

According to the given, f(x-a)+f(x+0.5a)=f(x-0.5a)+f(x), where x and a are rational. Likewise f(x-0.5a)+f(x+a)=f(x+0.5a)+f(x). Hence f(x+a)-f(x)= f(x)-f(x-a), namely 2f(x)=f(x-a)+f(x+a). Let f(0)=C, then consider F(x)=f(x)-C, where F(0)=0, 2F(x)=F(x-a)+F(x+a).

F(2x)=F(x)+[F(x)-F(0)]=2F(x), F(3x)=F(2x)+[F(2x)-F(x)]=3F(x). Easily, by induction, F(nx)=nF(x) for all integers k. Therefore, for nonzero integer m, (1/m)F(mx)=F(x) , namely F(x/m)=(1/m)F(x) Hence F(n/m)=(n/m)F(1). Let F(1)=k, we obtain F(x)=kx, where k is the slope of the linear functions, and f(x)=kx+C.

Page

Toolbox

Search

2015 USAMO Problems/Problem 4

Problem

Solution