Greatest common divisor of polynomials

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Discussion Overview

The discussion revolves around finding the greatest common divisor (gcd) of the polynomials \(x^4+1\) and \(x^2+x+1\). Participants explore the application of the Euclidean algorithm and consider different contexts, such as ordinary polynomials versus polynomials over integer modulo 2.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant initially claims that the gcd is \(x+1\) based on their application of the Euclidean division.
  • Another participant points out that the Euclidean algorithm was not fully completed, suggesting that further steps are necessary to determine the correct gcd.
  • A later reply acknowledges the oversight and concludes that the gcd is indeed \(1\) after completing the algorithm.
  • Another participant reiterates the initial claim and discusses the factorization of the polynomials, indicating that if considered as ordinary polynomials, they have no common factors, leading to a gcd of \(1\).
  • This participant also presents an alternative perspective, stating that if the polynomials are viewed as integer modulo 2 polynomials, their factorization leads to a gcd of \(1 + x + x^2\).

Areas of Agreement / Disagreement

There is no consensus on the gcd, as participants present competing views based on different interpretations of the polynomials. Some argue for \(1\) as the gcd, while others suggest \(1 + x + x^2\) in a different context.

Contextual Notes

The discussion highlights the dependence on the context in which the polynomials are considered, such as ordinary polynomials versus polynomials over integer modulo 2, which affects the outcome of the gcd.

evinda
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Hello! :cool:

I want to find the greatest common divisor of $x^4+1$ and $x^2+x+1$.
I applied the Euclidean division and found that $x^4+1=(x^2+x+1) \cdot (x^2-x)+(x+1)$.
So,isn't it like that: $gcd(x^4+1,x^2+x+1)=x+1$ ?
But.. in my textbook,the result is $1$! Which of the both results is right?? (Blush) :confused: (Thinking)
 
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You did not finish the Euclidean algorithm, which consists of several Euclidean divisions.
 
Evgeny.Makarov said:
You did not finish the Euclidean algorithm, which consists of several Euclidean divisions.

Oh,yes!You are right! (Giggle) I found now that the greatest common divisor is equal to $1$! Thank you! :)
 
evinda said:
Hello! :cool:

I want to find the greatest common divisor of $x^4+1$ and $x^2+x+1$.
I applied the Euclidean division and found that $x^4+1=(x^2+x+1) \cdot (x^2-x)+(x+1)$.
So,isn't it like that: $gcd(x^4+1,x^2+x+1)=x+1$ ?
But.. in my textbook,the result is $1$! Which of the both results is right?? (Blush) :confused: (Thinking)

If they are 'ordinary polynomials' then their factorisations are...

$\displaystyle 1 + x^{4} = (x - e^{i\ \frac{\pi}{4}})\ (x - e^{- i\ \frac{\pi}{4}})\ (x - e^{i\ \frac{3\ \pi}{4}})\ (x - e^{- i\ \frac{3\ \pi}{4}})$

$\displaystyle 1 + x + x^{2} = (x - e^{i\ \frac{2\ \pi}{3}})\ (x - e^{- i\ \frac{2\ \pi}{3}})$

... and the two polynomial have no common factors so that the g.c.d. is 1.

If they are 'integer modulo 2 polynomials' however then their factorisations are...$\displaystyle 1 + x^{4} = (1 + x)\ (1 + x)\ (1 + x + x^{2})$$\displaystyle 1 + x + x^{2} = 1 + x + x^{2}$ ... so that the g.c.d. is $\displaystyle 1 + x + x^{2}$...

Kind regards $\chi$ $\sigma$
 

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