1
$\begingroup$

This question already has an answer here:

Let $\mathbb F$ be any field $a \neq b$ be two elements of $\mathbb F$

Find the GCD of $ f(x) = x + a $ and $g(x) = x + b$. Also find the polynomials $s(x)$ and $t(x)$ such that $s(x)f(x) + t(x)g(x)$ equals the GCD.

My work so far:

$d(x) = s(x)f(x)+t(x)g(x)$

$d(x) = s(x)[x+a] + t(x)[x+b]$

$d(x) = x[s(x) + t(x)] + s(x)*a +t(x)*a$

Thoughts on how I can find the polynomials? Is there an explicit solution for the GCD $d(x)$?

$\endgroup$

marked as duplicate by Tom Oldfield, Andreas Caranti, Ross Millikan, user23500, vonbrand Apr 3 '13 at 21:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This question was asked and answered a day or so ago. $\endgroup$ – André Nicolas Apr 3 '13 at 21:19
  • $\begingroup$ Now that you know that, you could try to prove that the gcd of $X^n-1$ and $X^m-1$ is $X^k-1$ with $k$ the gcd of $m$ and $n$. $\endgroup$ – Julien Apr 3 '13 at 21:54
  • $\begingroup$ Interesting. Thank you for the suggestion. $\endgroup$ – RulesOfTheGame Apr 3 '13 at 22:10
0
$\begingroup$

if $a\neq b$, theyre coprime and $[(x+a)-(x+b)](a-b)^{-1}=1$

$\endgroup$
  • $\begingroup$ We know they are coprime because we can do a case such as a = 1 and b = 2? $\endgroup$ – RulesOfTheGame Apr 3 '13 at 22:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.