Determine whether or not two polynomials $P, Q$ with degree no less than 2018 and with integer coefficients exist such that $$P(Q(x))=3Q(P(x))+1$$for all real numbers $x$.
Problem
Source: 2018 FKMO Problem 5
Tags: algebra, FKMO, Korea
25.03.2018 08:46
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.
25.03.2018 08:49
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.
25.03.2018 08:50
I also tried this problem in test, but I can't know even is it exist or not. Is it exist?
25.03.2018 08:50
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!!
25.03.2018 09:26
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.
25.03.2018 09:31
Solution pls..........
25.03.2018 10:23
I'm really curious about cut score. $1,2,4$ are easy but $3,5,6$ are hard to solve.
25.03.2018 20:04
The thing is... 1,2,4 was too easy and 3,5,6 was too difficult... now everyone has the same score.
25.03.2018 20:41
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.
25.03.2018 21:08
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
26.03.2018 10:05
$P(x)=((6x+3)^n-1)/2$and $Q(x)=((6x+3)^{n-1}-1)/2$ is solution of this equation.
18.06.2018 14:21
One may well begin with $P(x) =2x+2x^2$ and $Q(x) = 1+3x$.
23.07.2018 17:40
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?
17.06.2019 04:53
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
22.06.2021 07:20
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.
22.06.2021 07:55
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.