2015 USAJMO Problems/Problem 4


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 $\boxed{f(x)=kx+c}$.