2011-04-14, 06:11 | #1 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3^{2}·17·31 Posts |
2LMs with exponent divisible by 13
I wanted to ask for help with algebra. Nobody had done a factorization in this manner before (a.f.a.i.k), so it could very well be that the answer is simply negative. Or maybe nobody wanted to deal with degree-6/2 polynomial pair before.
For Cunningham Aurifeuillians, 2,7N{L|M}, 2,5N{L|M} or 2,3N{L|M} are easily reduced to a sextic and quartics with a linear second poly. Now, the exercise is for 2,13N{L|M} cofactors. Example: 2,1846L. Here's the F_{24}(x) for it: gp> factor(2^26*x^52+1) %1 = [2*x^2 - 2*x + 1 1] [2*x^2 + 2*x + 1 1] [4096*x^24 - 4096*x^23 + 2048*x^22 - 1024*x^20 + 1024*x^19 - 512*x^18 + 256*x^16 - 256*x^15 + 128*x^14 - 64*x^12 + 32*x^10 - 32*x^9 + 16*x^8 - 8*x^6 + 8*x^5 - 4*x^4 + 2*x^2 - 2*x + 1 1] [4096*x^24 + 4096*x^23 + 2048*x^22 - 1024*x^20 - 1024*x^19 - 512*x^18 + 256*x^16 + 256*x^15 + 128*x^14 - 64*x^12 + 32*x^10 + 32*x^9 + 16*x^8 - 8*x^6 - 8*x^5 - 4*x^4 + 2*x^2 + 2*x + 1 1] The reciprocally-reduced (y=2*x+1/x) polynomial for one of the cofactors is f = y^12+2*y^11-22*y^10-44*y^9+172*y^8+344*y^7-560*y^6-1120*y^5+672*y^4+1344*y^3-224*y^2-448*y-64 y is rational e.g. with y=(2^71+1)/2^35, f=(2,1846L/2,142M)/2^420 (Footnote: I may have swapped L and M, above. But is there's a plan for L, there will be one for M, as well. Same goes for divisibility by 11.) Could there exist a suitable transformed pair of polynomials f_{6}(z) = 6-th degree poly(z) g_{2}(z) = 2-nd degree poly(z) ? Can anyone devise a clever plan for a reduction of f(y), or f(y)/y^6, or f(y)/y^4, where y is a root of a quadratic? Or maybe a step back is needed and the F_{24}(x) can be a resultant of a f_{6}(z) and g_{4}(z) ? What do you think? |
2011-04-14, 10:23 | #2 |
"Robert Gerbicz"
Oct 2005
Hungary
2715_{8} Posts |
There is no integer pair of polynomials for the problem: y=A*x^2+B*x+C+D/x+E/x^2
f(y)=x^12*(a6*y^6+a5*y^5+a4*y^4+a3*y^3+a2*y^2+a1*y+a0) for that f(y)=[4096*x^24 - 4096*x^23 + 2048*x^22 - 1024*x^20 + 1024*x^19 - 512*x^18 + 256*x^16 - 256*x^15 + 128*x^14 - 64*x^12 + 32*x^10 - 32*x^9 + 16*x^8 - 8*x^6 + 8*x^5 - 4*x^4 + 2*x^2 - 2*x + 1 1] Because polcoeff(f(y),1)=6*a6*E^5*D=-2 is impossible for integer a6,D,E. |
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