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Let $P$ and $Q$ be two polynomials with complex coefficients. It is known that $P(x^2+x+1)=Q(x^2-x+1)$. How can I prove that $P$ and $Q$ are constants?

I evaluated at various points to deduce $Q(3) = P(1) = Q(1) = P(3)$ but I don't know how to conclude from that.

darij grinberg
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Student12
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1 Answers1

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Set $x=0$ and you get $P(1)=Q(1)$

With $x=1-0=1$ you get $P(3)=Q(1)$ and $x=-1-0=-1$ gives $P(1)=Q(3)$

With $x=-1-1=-2$ you get $P(3)=Q(7)$ and with $x=1-(-1)=2$ there is $Q(3)=P(7)$

Notice that $x^2+x+1=(-1-x)^2+(-1-x)+1$ and $x^2-x+1=(1-x)^2-(1-x)+1$.

So you can use vieta jumping with the functional relation to show (for $n\in \mathbb Z$)$$\dots P(7)=Q(3)=P(1)=Q(1)=P(3)=Q(7)= \dots$$

But a polynomial $P$ which takes the same value $c$ at an infinite number of points is constant, since then $P-c$ has infinitely many roots hence $P-c = 0$.

To be more explicit $$P(n^2+n+1)=Q(n^2-n+1)=Q((1-n)^2-(1-n)+1)=P((1-n)^2+(1-n)+1)=$$$$[=P(n^2-3n-3)]=P((-1-(1-n))^2+(-1-(1-n))+1)=P((n-2)^2+(n-2)+1)$$ is enough.

Bill Dubuque
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Mark Bennet
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    I made a vieta-jumping tag a few years ago, I edited that into the question. There are just a few dozen questions with that tag, maybe I'm the only one that uses it. – Will Jagy Jan 03 '18 at 20:09
  • @WillJagy Thanks. I think that the original IMO question which really brought this to wider attention did identify a method which was not taught or well known, which is simple and which has wider application. Of course this opens the question as to how/whether one can identify simple relations between roots at higher degree which would enable similar elementary "jumping". Or is this all really Galois theory? – Mark Bennet Jan 03 '18 at 23:22
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    The best discussion I know is Hurwitz (1907), I put a good copy at http://zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf Oh, this was asked about at MO and I gave an answer: to me it is just quadratic forms, their automorphism groups, and inequalities, which are the only part that can sometimes be difficult https://mathoverflow.net/questions/289572/underlying-structure-behind-the-infamous-imo-1988-problem-6/289583#289583 – Will Jagy Jan 03 '18 at 23:31
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    @WillJagy Thanks for that - so this is a quadratic phenomenon (as the pairing would suggest). I might ask sometime whether there are generalisations. – Mark Bennet Jan 03 '18 at 23:47
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    I added some links (or course you can edit / reject as you like). The question is up for deletion by CRUDE though I appealed for mercy since it's quite instructive. It could use some reopen votes @Will too. Maybe someone can think of some "context" or better "effort" for the question. – Bill Dubuque Dec 19 '19 at 18:13
  • @BillDubuque Your improvements are always welcome. There is a tendency to try to close questions with real mathematical content on various grounds. I voted to re-open another one because the question revealed that the OP did not understand a basic concept. The people who closed it wanted the OP to state that in the question, but sometimes people who don't know stuff don't know they don't know it. I upvoted the question here and enjoyed writing the answer, because I thought it would help. If I enjoy writing the answer, I reckon the question is worth saving. – Mark Bennet Dec 19 '19 at 21:24