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.
Answer: any linear function $f$. These work.  $$ \begin{aligned} f(a)+f(a+3 d) & =f(a+d)+f(a+2 d) \\ f(a-d)+f(a+2 d) & =f(a)+f(a+d) \end{aligned} $$  which imply  $$ f(a-d)+f(a+3 d)=2 f(a+d) $$  Thus we conclude that for arbitrary $x$ and $y$ we have  $$ f(x)+f(y)=2 f\left(\frac{x+y}{2}\right) $$  thus $f$ satisfies Jensen's functional equation over $\mathbb{Q}$, so $f$ is linear.  $$ \begin{aligned} & f\left(\frac{2 n-1}{2 d}\right)+f\left(\frac{2 n+2}{2 d}\right)=f\left(\frac{2 n}{2 d}\right)+f\left(\frac{2 n+1}{2 d}\right) \\ & f\left(\frac{2 n-2}{2 d}\right)+f\left(\frac{2 n+1}{2 d}\right)=f\left(\frac{2 n-1}{2 d}\right)+f\left(\frac{2 n}{2 d}\right) \end{aligned} $$  Summing them and simplifying implies that  $$ f\left(\frac{n-1}{d}\right)+f\left(\frac{n+1}{d}\right)=2 f\left(\frac{n}{d}\right) $$  or equivalently  $$ f\left(\frac{n}{d}\right)-f\left(\frac{n-1}{d}\right)=f\left(\frac{n+1}{d}\right)-f\left(\frac{n}{d}\right) . $$  This implies that on the set of rational numbers with denominator dividing $d$, the function $f$ is linear.  In particular, we should have $f\left(\frac{n}{d}\right)=f(0)+\frac{n}{d} f(1)$ since $\frac{n}{d}, 0,1$ have denominators dividing $d$. This is the same as saying $f(q)=f(0)+q(f(1)-f(0))$ for any $q \in \mathbb{Q}$, which is what we wanted to prove.