What is the $GCD(2^{32}-2^{24}+2^{16}-2^8+1, 2^8+1)$?

Question

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

For to get $GCD(2^{32}-2^{24}+2^{16}-2^8+1, 2^8+1)$ I make the result that $(x^5+y^5) = (x+y)(x^4 - x^3y + x^2y^2-xy^3+y^4)$ and then, I choose $x = 2^8 $ and $y = 1$.

By the dupe the gcd $,(f(x),x+1) = (f(-1),x+1),,$ so specializing $,f(x) = x^4!-!x^3!+!x^2!-!x!+!1,\ x = 2^8\Rightarrow $ your gcd $= (4,x+1)=1,,$ by $x+1$ odd. $\ \ $ — Bill Dubuque, Mar 03 '24 at 00:09

score 5 · Answer 1 · answered Jul 12 '18 at 18:26

5

$x=2^8$ then $2^{32}-2^{24}+2^{16}-2^8+1=x^4-x^3+x^2-x+1$

Now, let's switch from numbers to polynomials $$x^4-x^3+x^2-x+1 = (x+1)Q(x)+5$$ Where $Q(x)$ is a polynomial with integer coefficients and you can easily see the equality by substituting $x=-1$

From this and Euclidean algorithm: $$ \text{gcd}(\frac{2^{40}+1}{2^8+1}, 2^8+1) = \text{gcd}(5, 2^8+1) = 1$$

answered Jul 12 '18 at 18:26

Jakobian

10,247

1

Good. Of course $2^8+1=257$, which is either well known to be prime, or verifiably prime with a few moments’ thought, so relatively prime to $5$. – Lubin Jul 12 '18 at 18:44
Mathematica yields $1$ instantly. – David G. Stork Jul 12 '18 at 18:49

score 2 · Accepted Answer · answered Jul 12 '18 at 18:33

2

Hint

$$2^8 \equiv -1 \pmod{2^8+1}$$

Therefore $$2^{32}-2^{24}+2^{16}-2^8+1 \equiv (-1)^4-(-1)^3+(-1)^2-(-1)+1 \pmod{2^8+1}$$

answered Jul 12 '18 at 18:33

N. S.

132,525

score 2 · Answer 3 · answered Jul 12 '18 at 19:28

\begin{array}{rrrrrrrrrrr} & & & 2^{24} &-2\cdot2^{16} &+3\cdot 2^8 &-4 \\ & &--- &--- &--- &--- &---\\ 2^8+1 &| &2^{32} &-2^{24} &+2^{16} &-2^8 &+1 \\ & &2^{32} &+2^{24} \\ & &--- &---\\ & & &-2\cdot2^{24} & +2^{16} \\ & & &-2\cdot2^{24} &-2\cdot 2^{16} \\ & & &--- &---\\ & & & &3\cdot 2^{16} & -2^8 \\ & & & &3\cdot 2^{16} &+3\cdot 2^8 \\ & & & &--- &---\\ & & & & &-4\cdot 2^8 & +1\\ & & & & &-4\cdot 2^8 & -4\\ & & & & &--- &---\\ & & & & & & +5 \end{array}

\begin{align} \gcd(2^{32}-2^{24}+2^{16}-2^8+1,2^8+1) &= \gcd(2^8+1, 5) \\ &= \gcd(257, 5) \\ &= 1 \end{align}

What is the $GCD(2^{32}-2^{24}+2^{16}-2^8+1, 2^8+1)$?

3 Answers3