2015 USAJMO 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 1

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 $n$ . 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 $f(x)=kx+C$ .

Solution 2

We have $f(x-3d)+f(x+3d)=f(x-d)+f(x+d)$ and $f(x)+f(x+3d)=f(x+d)+f(x+2d).$ Subtracting these two and rearranging gives $f(x-3d)+f(x+2d)=f(x)+f(x-d),$ and since $f(x+2d)=f(x+d)+f(x)-f(x-d)$ we get $f(x-3d)+f(x+d)=2f(x-d)$ from which we get $f(x-d)+f(x+d)=2f(x).$ Then we have $f(x)+f(y)=f(0)+f(x+y)=2f\left(\frac{x+y}{2}\right)$ . Setting $f(0)=c$ , we let $f(x)=g(x)+c$ to get $g(x)+g(y)=g(x+y)$ . This is Cauchy's functional equation, so it has solutions at $g(x)=kx$ , so the answer is $\boxed{f(x)=kx+c}$ .

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2015 USAJMO Problems/Problem 4

Problem

Solution 1

Solution 2