I found this question in a Russian Olympiad book and has left me completely stupefied. Though I have done questions regarding abstract concepts like the Pigeon Hole principle, but still because I am new to such concepts I have been unable to even start thinking about the solution.
The question goes like this:
Can you place $15$ integers around a circle such that sum of every $4$ consecutive numbers is either $1 $ or $ 3$?
The question puts no constraints over the distinctness of the integers and has been framed exactly in the same manner.
It is not possible. Proceed by contradiction, suppose we had such an arrangement,$\{a_n\}_{n=1}^{15}$. Considering all $15$ possible sums of $4$ consecutive values, we say that there are $n$ that add to $3$ and $m$ that add to $1$. Of course $$n+m=15$$
Let $\sum a_n$ be the sum of all the integers on your circle. Clearly, each integer appears in exactly $4$ sums. Thus $4\sum a_n$ is the sum of all the separate sums of $4$ consecutive integers. Thus $$4\sum a_n=3n+m$$
Subtracting our first equation from this one we see that $$2n=4\sum a_n-15$$ which is impossible (as $15$ is odd).