First $P(x) = x^2+3x+1$ and $Q(x) = 3x^2+9x+6$ satisfied the condition. And if $P(x)$ and $Q(x)$ satisfy the condition, either $P(Q(x))$ and $Q(x)$ or $P(x)$ and $Q(P(x))$ too. So it exists.

kjy1102 wrote: First P(x) = x^2+3x+1 Q(x) = 3x^2+9x+6 it satisfied the condition. And if P(x) and Q(x) satisfy the condition, P(Q(x)) and Q(x) too. So it exists. Degree must not less then 2018.

I also tried this problem in test, but I can't know even is it exist or not. Is it exist?

If P(x) and Q(x) works, so does P(Q(x)),Q(x) and P(x),Q(P(x)), so we can increase the degrees. Great solution anyway!!

Kinda like this problem https://artofproblemsolving.com/community/c6h546349p3161954 in a sense that nearly everybody guessed that the answer was negative, which was not the case. Respect to everyone who solved this in the contest.

Solution pls..........

I'm really curious about cut score. $1,2,4$ are easy but $3,5,6$ are hard to solve.

The thing is... 1,2,4 was too easy and 3,5,6 was too difficult... now everyone has the same score.

cloneofsimo wrote: The thing is... 1,2,4 was too easy and 3,5,6 was too difficult... now everyone has the same score. 5 wasn't that difficult, it was just that everyone expected the answer to be 'NO' because of the similar question almost every contestant saw a month ago.

that happened to me one time in a similar problem , well when you believe the answer is no you are dead , dead....it's really a big troll for contestants

$P(x)=((6x+3)^n-1)/2$and $Q(x)=((6x+3)^{n-1}-1)/2$ is solution of this equation.

One may well begin with $P(x) =2x+2x^2$ and $Q(x) = 1+3x$.

jujunghun wrote: $P(x)=((6x+3)^n-1)/2$and $Q(x)=((6x+3)^{n-1}-1)/2$ is solution of this equation. I think this would be the answer. How did you determine it though?

I found a different construction $P(x) = \frac{3}{2}[(2x + 1)^{n+1}-1], Q(x) = \frac{3}{2}[(2x + 1)^n-1]$, which came from $3x[(2x+1)^n + (2x+1)^{n-1} + ... + 1]$. I guessed this construction from the small cases I did, and I feel that if (P, Q) works then (something, P) probably works, so I tried to inductively construct a pair of polynomials (though I was dumb not to do the P(Q(x)) solution). How to guess that the answer is yes? Well if the answer is no, I doubt the question will include "with degree no less than 2018", and also it is possible for smaller degrees XD

Solved with Alan Bu, Christopher Qiu, David Dong, Elliott, Liu, Isaac Zhu, Jeffrey Chen, Kevin Wu, and Ryan Yang. Yes: First \((P(x),Q(x))=(x^2+3x+1,3x+3)\) works. Next \((P,Q)\) works implies \((P\circ Q,Q)\) and \((P,Q\circ P)\) work.

If the pair $(P,Q)$ fits, then $(P\circ Q,Q)$ and $(P,Q\circ P)$ also work. After some testing, we find that $(2x+2x^2,1+3x)$ works, and we can always increase the degree of $P$ and $Q$ with the above method, hence the claim.