Before trying the problem "note your opinion as to whether the observed pattern is known to continue, known not to continue, or not known at all."
Lets factor polynomials of the form x^n1. Starting with 1 the following list can be generated:
n=1: (x1)
n=2: (x1)*(x+1)
n=3: (x1)*(x^2+x+1)
n=4: (x1)*(x+1)*(x^2+1)
n=5: (x1)*(x^4+x^3+x^2+x+1)
n=6: (x1)*(x+1)*(x^2+x+1)*(x^2x+1)
One thing to notice is that each line has exactly one polynomial factor not seen earlier in the list:
n=1: x1
n=2: x+1
n=3: x^2+x+1
n=4: x^2+1
n=5: x^4+x^3+x^2+x+1
n=6: x^2x+1
Does each new factorization always produce exactly one new polynomial factor?
Looking more closely you may see that all the coefficients are 1, 0, or 1. Does this continue to be the case for all factors?
(In reply to
re: Answering a different question... by Math Man)
Using the method I sketched out:
x^(1051) = (x^71)(x^98+x^91+x^84+x^77+x^70+x^63+x^56+x^49+x^42+x^35+x^28+x^21+x^14+x^7+1)[1]
but
(x^98+x^91+x^84+x^77+x^70+x^63+x^56+x^49+x^42+x^35+x^28+x^21+x^14+x^7+1)/(x^14+x^13+x^12+x^11+x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1) =
(x^84x^83+x^77x^76+x^70x^68+x^63x^61+x^56x^53+x^49x^46+x^42x^38+x^35x^31+x^28x^23+x^21x^16+x^14x^8+x^7x+1) [2] (the invariant)
And as I pointed out, it is possible on occasion both to further factorise the invariant, and to factorise it in a way that produces a larger coefficient, depending on how one chooses to do it (compare the example I gave of x^61 = (x1)(x^2x+1)(x^3+2x^2+2x+1):
(x^84x^83+x^77x^76+x^70x^68+x^63x^61+x^56x^53+x^49x^46+x^42x^38+x^35x^31+x^28x^23+x^21x^16+x^14x^8+x^7x+1)
=
(x^24x^23+x^19x^18+x^17x^16+x^14x^13+x^12x^11+x^10x^8+x^7x^6+x^5x+1)(x^60x^55+x^45x^40+x^30x^20+x^15x^5+1),
and
x^60x^55+x^45x^40+x^30x^20+x^15x^5+1)
=
(x^24x^23+x^19x^18+x^17x^16+x^14x^13+x^12x^11+x^10x^8+x^7x^6+x^5x+1)(x^48+x^47+x^46x^43x^422x^41x^40x^39+x^36+x^35+x^34+x^33+x^32+x^31x^28x^26x^24x^22x^20+x^17+x^16+x^15+x^14+x^13+x^12x^9x^82x^7x^6x^5+x^2+x+1)
where the terms in x^41 and x^7 have coefficients of 2.
The point I was trying to make is that in order to make the sort of determination called for by the puzzle, a consistent  comparing like with like  approach is needed, which is why I believe it more instructive to focus on the invariant than its subsequent possible factorisations.
Edited on February 13, 2019, 10:52 pm

Posted by broll
on 20190213 22:33:13 