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March 21st, 2019, 09:42 PM   #1
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Lightbulb Polynomials divisibilty problem

If P(x) and Q(x) are two polynomials such that P(x) | P(Q(x)), what are the restrictions for Q such that the statement P (x) = 0 => Q(x) = x to be true (I was thinking the restriction has to be Q to be strictly monotone on Im(Q), but I’m not quite sure)?
If the left to right implication is true, furthermore, can someone please help me prove whether the inverse implication ( If P(x) and Q(x) are two polynomials such that P(x) | P(Q(x)), then Q (x) = x => P (x) = 0 ) holds or is actually false?
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March 22nd, 2019, 06:00 AM   #2
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If $Q(x)=x$ then $P(Q(x))=P(x)$, which is obviously divisible by $P(x)$ regardless of anything else.

Last edited by v8archie; March 22nd, 2019 at 06:03 AM.
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March 22nd, 2019, 06:31 AM   #3
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I agree, however I think you didn’t exactly understood my question. We are given two polynomials, P(X) and Q(X), we know that P(x) | P (Q(x)), but we need to find the restriction for Q(x) such that P(x)=0 to imply that Q(x)=x ( normally, if P(x)|P(Q(x)) and P(x)=0, then Q(x) must be one of the roots of P(x), but not necessarily x. We need to find the restriction for Q such that the only possibility to actually be Q(x)=x). If you allow me to rephrase it a bit differently:

Let P(X),Q(X) belong to R[X] such that P(X) | P(Q(X)) and let’s consider the sets: A = { x in R | P(x) = 0} , B = {x in R | Q(x) = x}. I’m trying to find a necessary condition for Q such that A = B.
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