p(x)=x^{101}+x^{94}+x^{57}+x^{33}1
m(x)=ax^{2}+bx+c
g(x)=g1x^{99}+g2x^{98}+g3x^{97}+...+g99x + g100
Obviously: p(x)=m(x)*g(x)=
(ax^{2}+bx+c)*(g1x^{99}+g2x^{98}+g3x^{97}+...+g99x + g100)
We try with integers coefficients for a, b, c, and gi
c*g100=1 => c=1 g100=1 or the other way
b*g100+c*g99=0 =>g99=b
Going ahead it's clear that gi coefficients are always growing in complexity. But if a=b=1 then all the gi coefficients are 1 or 1 or 0 and the sequence (1,10) repeats
Then we have that with quadratic polynomial x^{2}+x+1 and with gi coefficients (1, 1, 0, 1, 1, 0, 1....) we are able to get expresions like
m(x)*q(x)=x^{101}+x+1
Now we need to introduce the other potencies and to eliminate the x and change the sign of 1
m(x)*g(x)=x^{94}x (when we limit gi to 93)
m(x)*g(x)=x^{57}1 (when we limit gi to 56)
m(x)*g(x)=x^{33}1 (when we limit gi to 32)
So the sum of the four polynomials is p(x)
p(x)=(x^{2}+x+1)*(g(x)
Coefficients of g(x) are= 1, 1, 0 g100>gi>g94
Coefficients of g(x) are= 1, 0, 1 g95>gi>g57
Coefficients of g(x) are= 0, 0, 0 g58>gi>g32
Coefficients of g(x) are= 1, 1, 0 g33>gi

Posted by armando
on 20190415 10:25:36 