P^n-1|Q^n-1
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From: blang on 29 Jun 2010 05:09 Hi :) Let P and Q in IR[X] (with deg(P)>0 and deg(Q)>0) such that : for all n in IN*, P^n-1|Q^n-1. Is it true that we can find k in IN* such that Q=P^k ? Thanks !
From: William Elliot on 29 Jun 2010 23:46 On Tue, 29 Jun 2010, blang wrote: > Let P and Q in IR[X] (with deg(P)>0 and deg(Q)>0) such that : for all n > in IN*, P^n-1|Q^n-1. Do you mean p^(n-1) | q^(n-1) or (p^n - 1) | (q^n - 1) ? > Is it true that we can find k in IN* such that Q=P^k ? > No or I don't know.
From: blang on 29 Jun 2010 22:32 (P^n-1) | (Q^n-1) , of course ! >Do you mean >p^(n-1) | q^(n-1) >or >(p^n - 1) | (q^n - 1) ?
From: achille on 30 Jun 2010 03:03 On Jun 30, 2:32 pm, blang <bl...(a)laposte.net> wrote: > (P^n-1) | (Q^n-1) , of course ! > > >Do you mean > >p^(n-1) | q^(n-1) > >or > >(p^n - 1) | (q^n - 1) ? It is true for the special case p(z) = z. When p(z) = z, the condition p^n - 1 | q^n - 1 for all n => |q(z)| = 1 for any complex z satisfies |z| = 1. Since p \in R(z), this implies on the unit circle, q(z) q(1/z) = q(z)q(z*) = q(z)(q(z))* = |q(z)|^2 = 1. It is then easy to check q(z) must has the form z^k. No idea about the general case :-(
From: William Elliot on 30 Jun 2010 03:32
On Wed, 30 Jun 2010, blang wrote: > (P^n-1) | (Q^n-1) , of course ! > What happen to the problem statement? With it being removed from your reply, there's nothing to comment other than does p^n-1 mean p^(n-1) or p^n - 1, ie (p^n) - 1? >> Do you mean >> p^(n-1) | q^(n-1) >> or >> (p^n - 1) | (q^n - 1) ? Please include essential content _within_ your reply http://oakroadsystems.com/genl/unice.htm ---- |