What is the GDC(2^32-2^24+2^16-2^8+1, 2^8+1)GDC(232224+21628+1,28+1)?

My question is:
GCD(\frac{2^40+1}{2^8+1}, 2^8+1)GCD(240+128+1,28+1), but I'm stuck here:
GDC(2^32-2^24+2^16-2^8+1, 2^8+1)GDC(232224+21628+1,28+1). How I resolve this?

For to get GDC(2^32-2^24+2^16-2^8+1, 2^8+1)GDC(232224+21628+1,28+1) I made about the result that
(x^5+y^5) = (x+y)(x^4 - x^3y + x^2y^2-xy^3+y^4)(x5+y5)=(x+y)(x4x3y+x2y2xy3+y4) and then, I choose x = 2^8 x=28 and y = 1y=1.

1 Answer
Jul 14, 2018

The greatest common divisor of 2^32-2^24+2^16-2^8+1232224+21628+1 and 2^8+128+1 is 11

Explanation:

Note that:

257 = 2^8+1 = 2^(2^3)+1257=28+1=223+1

is a prime number - in fact one of the few known Fermat prime numbers.

So the only possible common factors of 2^8+128+1 and 2^32-2^24+2^16-2^8+1232224+21628+1 are 11 and 257257.

However, as you have noted in the question:

2^32-2^24+2^16-2^8+1 = (2^40+1)/(2^8+1)232224+21628+1=240+128+1

is of the form:

x^4-x^3y+x^2y^2-xy^3+y^4 = (x^5+y^5)/(x+y)x4x3y+x2y2xy3+y4=x5+y5x+y

The one factor (x+y) = 2^8+1(x+y)=28+1 of 2^40+1240+1 corresponds to the real fifth root of unity and (x+y)(x+y) is not automatically a factor of the remaining quartic x^4-x^3y+x^2y^2-xy^3+y^4x4x3y+x2y2xy3+y4 whose other linear factors are all non-real complex.

We can manually divide x^4-x^3y+x^2y^2-xy^3+y^4x4x3y+x2y2xy3+y4 by x+yx+y to get a polynomial remainder and then substitute x=2^8x=28 and y=1y=1 to check that this is not a special case...

x^4-x^3y+x^2y^2-xy^3+y^4 = (x+y)(x^3-2x^2y+3xy^2-4y^3)+5y^4x4x3y+x2y2xy3+y4=(x+y)(x32x2y+3xy24y3)+5y4

So the remainder is:

5y^4 = 5(color(blue)(1))^4 = 55y4=5(1)4=5

Since the remainder is non-zero, 2^32-2^24+2^16-2^8+1232224+21628+1 and 2^8+128+1 have no common factor larger than 11.