p(x)=x101+x94+x57+x33-1
m(x)=ax2+bx+c
g(x)=g1x99+g2x98+g3x97+...+g99x + g100
Obviously: p(x)=m(x)*g(x)=
(ax2+bx+c)*(g1x99+g2x98+g3x97+...+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,-1-0) repeats
Then we have that with quadratic polynomial x2+x+1 and with gi coefficients (1, -1, 0, 1, -1, 0, 1....) we are able to get expresions like
m(x)*q(x)=x101+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)=x94-x (when we limit gi to 93)
m(x)*g(x)=x57-1 (when we limit gi to 56)
m(x)*g(x)=x33-1 (when we limit gi to 32)
So the sum of the four polynomials is p(x)
p(x)=(x2+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
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Posted by armando
on 2019-04-15 10:25:36 |
Solution (not very orthodox method)
