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I can prove that $X^a - 1 \mid X^b - 1$ if $a\mid b$, and even that the integer division of $a$ by $b$ translates in some sense into a division of $X^a$ by $X^b$. However, I do not know how to deduce $$\gcd(X^a-1, X^b-1) = X^{\gcd(a,b)}-1$$

I would like to know both formally why it can be proved, but also if there is something deeper to be understood (e.g. if there is a hidden notion of "morphism preserving the integer division" or so)

Bill Dubuque
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TheStudent
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1 Answers1

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You can show that for $m\ge n$ we have $$ \gcd(X^m - 1, X^n - 1) = \gcd(X^{m-n}-1, X^n-1), \tag{$\star$} $$ which mimics $\gcd(m,n)=\gcd(m-n,n)$ exactly and allows you to perform the Euclidean algorithm on $m$ and $n$ instead of $X^m-1$ and $X^n-1$.

Here's the proof of $(\star)$: \begin{align} \gcd(\color{red}{X^m - 1}, \color{blue}{X^n - 1}) &= \gcd(\color{red}{X^m-1}-X^{m-n}(\color{blue}{X^n-1}), \color{blue}{X^n-1}) \\ &= \gcd(\color{green}{X^{m-n}-1},\color{blue}{X^n-1}). \end{align}

Thus, using the same steps as in the Euclidean algorithm for $\gcd(m,n)$ you end up with $$ \gcd(X^m-1,X^n-1) = \gcd(X^{\gcd(m,n)}-1,\underbrace{X^0-1}_0) = X^{\gcd(m,n)}-1. $$

Christoph
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