HMMT-Nov 2016 guts, Problem 22

Let the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ take only integer inputs and have integer outputs. For any integers $x$ and $y, f$ satisfies \[ f(x)+f(y)=f(x+1)+f(y-1) \] If $f(2016)=6102$ and $f(6102)=2016$, what is $f(1)$ ?

Solution: We have \[ f(x+1)=f(x)+f(y)-f(y-1) \] If $y$ is fixed, we have \[ f(x+1)=f(x)+\text { constant } \] implying $f$ is linear. Using our two points, then, we get $f(x)=8118-x$, so $f(1)=8117$ $\fbox{8117}$.

Answer: 8117