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Old 2009-09-16, 17:24   #6
fivemack
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Feb 2006
Cambridge, England

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Quote:
Originally Posted by grandpascorpion View Post
Thank you both for your feedback.

I wonder if a better tack would be to check if one (or more) of these polynomials (in three variables: n,d and k) can be factored into two smaller polynomials say g(n,d,k) and h(n,d,k).
This is trivial with magma:

Code:
P<n,d,k>:=PolynomialRing(Rationals(),3);
F:=(4*n^6 - 4*d*n^5 - 3*d^2*n^4 + 6*d^3*n^3 - 4*d^4*n^2)*k^4 +
(8*n^7 + 8*d*n^6 - 26*d^2*n^5 + 10*d^3*n^4 + 6*d^4*n^3 - 12*d^5*n^2)*k^3 +
(4*n^8 + 28*d*n^7 - 23*d^2*n^6 - 32*d^3*n^5 + 42*d^4*n^4 - 24*d^5*n^3 - 8*d^6*n^2)*k^2 +
(16*d*n^8+ 16*d^2*n^7 - 52*d^3*n^6 + 20*d^4*n^5 + 12*d^5*n^4 - 24*d^6*n^3)*k +
(16*d^2*n^8 - 16*d^3*n^7 - 12*d^4*n^6 + 24*d^5*n^5 - 16*d^6*n^4);
Factorisation(F);
but the factors are uninteresting:

Code:
[
    <d + 1/2*k, 1>,
    <n, 2>,
    <n + k, 1>,
    <n^5*d + 1/2*n^5*k - n^4*d^2 + 1/2*n^4*d*k + 1/2*n^4*k^2 - 3/4*n^3*d^3 - 
        15/8*n^3*d^2*k - 1/2*n^3*d*k^2 + 3/2*n^2*d^4 + 5/4*n^2*d^3*k - 
        3/8*n^2*d^2*k^2 - n*d^5 - 1/4*n*d^4*k + 3/4*n*d^3*k^2 - 1/2*d^5*k - 
        1/2*d^4*k^2, 1>
]

Last fiddled with by fivemack on 2009-09-16 at 17:25
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