Problem:
Find all functions f:Q→Q such that
f(x)+f(t)=f(y)+f(z)
for all rational numbers x<y<z<t that form an arithmetic progression. ( Q is the set of all rational numbers.)
Solution:
Choose any n∈Z,t∈Q. Applying the condition for nt,(n+1)t,(n+2)t,(n+3)t yields
f((n+3)t)−f((n+2)t)=f((n+1)t)−f(nt)
and similarly
f((n+4)t)−f((n+3)t)=f((n+2)t)−f((n+1)t)
Adding the two yields
f((n+4)t)−f((n+2)t)=f((n+2)t)−f(nt)
in particular f(2kt+2t)−f(2kt) is the same for all k∈Z, which means f is linear on 2t⋅Z. Since Q is a nested union of such sets, f is linear and all linear functions work.
The problems on this page are the property of the MAA's American Mathematics Competitions