Divisibility of polynomials of several variables
in [Math]
|
Prev: heavy water in comets is already proof Re: Additive Creation; Dirac's new radioactivities Chapt 5 #169; ATOM TOTALITY
Next: Quick Question
From: leox on 15 Jun 2010 17:47 Will divide the polynomial (1-x^i y^j z^k) by the polynomial (1-x z) (1-yz^m) exactly with no remainder for some set of natural numbers i,j,k,m>=0? If yes, find such a value of i,j,k,m.
From: Robert Israel on 15 Jun 2010 20:07 leox <leonid.uk(a)gmail.com> writes: > Will divide the polynomial (1-x^i y^j z^k) by the polynomial (1-x > z) (1-yz^m) exactly with no remainder for some set of natural > numbers i,j,k,m>=0? If yes, find such a value of i,j,k,m. If polynomial A is divisible by polynomial B (i.e. there is a polynomial C such that A = B*C), then A = 0 whenever B = 0. You want A(x,y,z) = 1 - x^i y^j z^k to be 0 when z = 1/x. But A(x,y,1/x) = 1 - x^(i-k) y^j, so you need j=0 and i=k. Similarly, A(x,z^(-m),z) = 0 = 1 - x^i z^(k-jm), so i=0 and k=jm. Putting these together, the only solution is i=0,j=0,k=0. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: leox on 16 Jun 2010 01:46
On 16 ЧеÑ, 03:07, Robert Israel <isr...(a)math.MyUniversitysInitials.ca> wrote: Thank you! |