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Finding Quadratic (Posted on 2019-04-12) Difficulty: 3 of 5
Find a quadratic polynomial which is a factor of x101+x94+x57+x33-1

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution (not very orthodox method) | Comment 1 of 2
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

  Posted by armando on 2019-04-15 10:25:36
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