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 $F(0)=0$ and $2F(x)=F(x-a)+F(x+a)$ . We have:

$F(2x)=F(x)+[F(x)-F(0)]=2F(x)$ $F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)$ By induction, $F(nx)=nF(x)$ for all in.tegers $k$ . Therefore, for nonzero integer $m$ , $\frac{1}{m}F(mx)=F(x)$ , namely $F\left(\frac{x}{m}\right)=\left(\frac{1}{m}\right)F(x)$ . Hence $F\left(\frac{n}{m}\right)=\left(\frac{n}{m}\right)F(1)$ . Letting $F(1)=k$ , we obtain $F(x)=kx$ , where $k$ is the slope of the linear functions, and $f(x)=kx+C$ .

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2015 USAMO Problems/Problem 4

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